Finding Partial Sum for Harmonic Series [closed]

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The harmonic series $1+frac12+frac13+cdots$ does not have a sum. How many terms $n$ are required for the $nth$ partial sum $S_n$ to be greater than (a) $2$, (b) $3$, (c) $5$?

I am not sure on how to find the sum but I know it sums up to infinity.

sequences-and-series harmonic-numbers

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edited Aug 2 at 1:47

Simply Beautiful Art

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asked Jul 28 at 0:01

user139289

144

closed as off-topic by John Ma, Adrian Keister, Taroccoesbrocco, amWhy, Xander Henderson Aug 2 at 1:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Adrian Keister, Taroccoesbrocco, amWhy, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Do not try to find a formula. Just try adding some of the first few terms. Find when your sum surpasses $2$.
  – GEdgar
  Jul 28 at 0:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The partial sums of the harmonic series are Riemann sums, with a uniform partition, for the integral $int_1^nfracdxx=ln(n)$. Therefore, the solution of $ln(n)=k$, $n=e^k$, gives you an estimate of how many terms to sum to get roughly $k$.
  – user578878
  Jul 28 at 0:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GEdgar, I did try that. But how do you find n for each case?
  – user139289
  Jul 28 at 0:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a famous proof of the divergence of the series. The proof compares this series with another series, containing powers of $2$. You can use the latter to get at least a bound.
  – Cristhian Grundmann
  Jul 28 at 0:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  why are you so reluctant to take @GEdgar's advice and do it by hand? 2 and 3 are both doable by hand, 5 to about 15 can easily be done with a simple computer program (e.g. for 15 you need almost 2 million terms). Anything bigger than that is probably best done approximately using Euler's estimate.
  – NickD
  Jul 28 at 2:15
  

  
  
  
  

 | 
show 5 more comments

up vote
0
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1

The harmonic series $1+frac12+frac13+cdots$ does not have a sum. How many terms $n$ are required for the $nth$ partial sum $S_n$ to be greater than (a) $2$, (b) $3$, (c) $5$?

I am not sure on how to find the sum but I know it sums up to infinity.

sequences-and-series harmonic-numbers

share|cite|improve this question

edited Aug 2 at 1:47

Simply Beautiful Art

49.1k572169

asked Jul 28 at 0:01

user139289

144

closed as off-topic by John Ma, Adrian Keister, Taroccoesbrocco, amWhy, Xander Henderson Aug 2 at 1:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Adrian Keister, Taroccoesbrocco, amWhy, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Do not try to find a formula. Just try adding some of the first few terms. Find when your sum surpasses $2$.
  – GEdgar
  Jul 28 at 0:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The partial sums of the harmonic series are Riemann sums, with a uniform partition, for the integral $int_1^nfracdxx=ln(n)$. Therefore, the solution of $ln(n)=k$, $n=e^k$, gives you an estimate of how many terms to sum to get roughly $k$.
  – user578878
  Jul 28 at 0:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GEdgar, I did try that. But how do you find n for each case?
  – user139289
  Jul 28 at 0:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a famous proof of the divergence of the series. The proof compares this series with another series, containing powers of $2$. You can use the latter to get at least a bound.
  – Cristhian Grundmann
  Jul 28 at 0:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  why are you so reluctant to take @GEdgar's advice and do it by hand? 2 and 3 are both doable by hand, 5 to about 15 can easily be done with a simple computer program (e.g. for 15 you need almost 2 million terms). Anything bigger than that is probably best done approximately using Euler's estimate.
  – NickD
  Jul 28 at 2:15
  

  
  
  
  

 | 
show 5 more comments

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

The harmonic series $1+frac12+frac13+cdots$ does not have a sum. How many terms $n$ are required for the $nth$ partial sum $S_n$ to be greater than (a) $2$, (b) $3$, (c) $5$?

I am not sure on how to find the sum but I know it sums up to infinity.

sequences-and-series harmonic-numbers

share|cite|improve this question

edited Aug 2 at 1:47

Simply Beautiful Art

49.1k572169

asked Jul 28 at 0:01

user139289

144

The harmonic series $1+frac12+frac13+cdots$ does not have a sum. How many terms $n$ are required for the $nth$ partial sum $S_n$ to be greater than (a) $2$, (b) $3$, (c) $5$?

I am not sure on how to find the sum but I know it sums up to infinity.

sequences-and-series harmonic-numbers

share|cite|improve this question

edited Aug 2 at 1:47

Simply Beautiful Art

49.1k572169

asked Jul 28 at 0:01

user139289

144

share|cite|improve this question

share|cite|improve this question

edited Aug 2 at 1:47

Simply Beautiful Art

49.1k572169

edited Aug 2 at 1:47

Simply Beautiful Art

49.1k572169

edited Aug 2 at 1:47

Simply Beautiful Art

49.1k572169

49.1k572169

asked Jul 28 at 0:01

user139289

144

asked Jul 28 at 0:01

user139289

144

asked Jul 28 at 0:01

user139289

144

144

closed as off-topic by John Ma, Adrian Keister, Taroccoesbrocco, amWhy, Xander Henderson Aug 2 at 1:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Adrian Keister, Taroccoesbrocco, amWhy, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by John Ma, Adrian Keister, Taroccoesbrocco, amWhy, Xander Henderson Aug 2 at 1:08

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John Ma, Adrian Keister, Taroccoesbrocco, amWhy, Xander Henderson

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Do not try to find a formula. Just try adding some of the first few terms. Find when your sum surpasses $2$.
  – GEdgar
  Jul 28 at 0:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The partial sums of the harmonic series are Riemann sums, with a uniform partition, for the integral $int_1^nfracdxx=ln(n)$. Therefore, the solution of $ln(n)=k$, $n=e^k$, gives you an estimate of how many terms to sum to get roughly $k$.
  – user578878
  Jul 28 at 0:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GEdgar, I did try that. But how do you find n for each case?
  – user139289
  Jul 28 at 0:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a famous proof of the divergence of the series. The proof compares this series with another series, containing powers of $2$. You can use the latter to get at least a bound.
  – Cristhian Grundmann
  Jul 28 at 0:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  why are you so reluctant to take @GEdgar's advice and do it by hand? 2 and 3 are both doable by hand, 5 to about 15 can easily be done with a simple computer program (e.g. for 15 you need almost 2 million terms). Anything bigger than that is probably best done approximately using Euler's estimate.
  – NickD
  Jul 28 at 2:15
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Do not try to find a formula. Just try adding some of the first few terms. Find when your sum surpasses $2$.
  – GEdgar
  Jul 28 at 0:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  The partial sums of the harmonic series are Riemann sums, with a uniform partition, for the integral $int_1^nfracdxx=ln(n)$. Therefore, the solution of $ln(n)=k$, $n=e^k$, gives you an estimate of how many terms to sum to get roughly $k$.
  – user578878
  Jul 28 at 0:07
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @GEdgar, I did try that. But how do you find n for each case?
  – user139289
  Jul 28 at 0:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  There is a famous proof of the divergence of the series. The proof compares this series with another series, containing powers of $2$. You can use the latter to get at least a bound.
  – Cristhian Grundmann
  Jul 28 at 0:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  why are you so reluctant to take @GEdgar's advice and do it by hand? 2 and 3 are both doable by hand, 5 to about 15 can easily be done with a simple computer program (e.g. for 15 you need almost 2 million terms). Anything bigger than that is probably best done approximately using Euler's estimate.
  – NickD
  Jul 28 at 2:15
  

  
  
  
  

2

2

Do not try to find a formula. Just try adding some of the first few terms. Find when your sum surpasses $2$.
– GEdgar
Jul 28 at 0:05

Do not try to find a formula. Just try adding some of the first few terms. Find when your sum surpasses $2$.
– GEdgar
Jul 28 at 0:05

The partial sums of the harmonic series are Riemann sums, with a uniform partition, for the integral $int_1^nfracdxx=ln(n)$. Therefore, the solution of $ln(n)=k$, $n=e^k$, gives you an estimate of how many terms to sum to get roughly $k$.
– user578878
Jul 28 at 0:07

The partial sums of the harmonic series are Riemann sums, with a uniform partition, for the integral $int_1^nfracdxx=ln(n)$. Therefore, the solution of $ln(n)=k$, $n=e^k$, gives you an estimate of how many terms to sum to get roughly $k$.
– user578878
Jul 28 at 0:07

@GEdgar, I did try that. But how do you find n for each case?
– user139289
Jul 28 at 0:08

@GEdgar, I did try that. But how do you find n for each case?
– user139289
Jul 28 at 0:08

1

1

There is a famous proof of the divergence of the series. The proof compares this series with another series, containing powers of $2$. You can use the latter to get at least a bound.
– Cristhian Grundmann
Jul 28 at 0:10

There is a famous proof of the divergence of the series. The proof compares this series with another series, containing powers of $2$. You can use the latter to get at least a bound.
– Cristhian Grundmann
Jul 28 at 0:10

1

1

why are you so reluctant to take @GEdgar's advice and do it by hand? 2 and 3 are both doable by hand, 5 to about 15 can easily be done with a simple computer program (e.g. for 15 you need almost 2 million terms). Anything bigger than that is probably best done approximately using Euler's estimate.
– NickD
Jul 28 at 2:15

why are you so reluctant to take @GEdgar's advice and do it by hand? 2 and 3 are both doable by hand, 5 to about 15 can easily be done with a simple computer program (e.g. for 15 you need almost 2 million terms). Anything bigger than that is probably best done approximately using Euler's estimate.
– NickD
Jul 28 at 2:15

 | 
show 5 more comments

1 Answer
1

active

oldest

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0
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accepted

The minimum number of terms needed so that harmonic series is at least $n$, can be found via oeis.org/A004080

The first few terms are:

1, 4, 11, 31, 83, 227, 616, 1674, 4550, 12367, 33617, 91380, 248397, 675214, 1835421, 4989191, 13562027, 36865412, 100210581, 272400600, 740461601, 2012783315, 5471312310, 14872568831, 40427833596, 109894245429, 298723530401, 812014744422

Thus,

The number of terms required for the sum to exceed 2 is 4.

The number of terms required for the sum to exceed 3 is 11.

The number of terms required for the sum to exceed 5 is 83.

Interestingly, after 2.5×10^8 terms, the sum is still less than 20. Furthermore, to achieve a sum greater than 100, the number of terms required is 15092688622113788323693563264538101449859497

On a related point, the number of terms needed to equal or exceed $10^n$ are given in the sequence (OEIS A096618).

The first three terms are:
12367, 15092688622113788323693563264538101449859497 and $1.1×10^434$, .

share|cite|improve this answer

edited Jul 28 at 2:33

answered Jul 28 at 2:23

Martin Roberts

1,204318

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

The minimum number of terms needed so that harmonic series is at least $n$, can be found via oeis.org/A004080

The first few terms are:

1, 4, 11, 31, 83, 227, 616, 1674, 4550, 12367, 33617, 91380, 248397, 675214, 1835421, 4989191, 13562027, 36865412, 100210581, 272400600, 740461601, 2012783315, 5471312310, 14872568831, 40427833596, 109894245429, 298723530401, 812014744422

Thus,

The number of terms required for the sum to exceed 2 is 4.

The number of terms required for the sum to exceed 3 is 11.

The number of terms required for the sum to exceed 5 is 83.

Interestingly, after 2.5×10^8 terms, the sum is still less than 20. Furthermore, to achieve a sum greater than 100, the number of terms required is 15092688622113788323693563264538101449859497

On a related point, the number of terms needed to equal or exceed $10^n$ are given in the sequence (OEIS A096618).

The first three terms are:
12367, 15092688622113788323693563264538101449859497 and $1.1×10^434$, .

share|cite|improve this answer

edited Jul 28 at 2:33

answered Jul 28 at 2:23

Martin Roberts

1,204318

add a comment | 

up vote
0
down vote



accepted

The minimum number of terms needed so that harmonic series is at least $n$, can be found via oeis.org/A004080

The first few terms are:

1, 4, 11, 31, 83, 227, 616, 1674, 4550, 12367, 33617, 91380, 248397, 675214, 1835421, 4989191, 13562027, 36865412, 100210581, 272400600, 740461601, 2012783315, 5471312310, 14872568831, 40427833596, 109894245429, 298723530401, 812014744422

Thus,

The number of terms required for the sum to exceed 2 is 4.

The number of terms required for the sum to exceed 3 is 11.

The number of terms required for the sum to exceed 5 is 83.

Interestingly, after 2.5×10^8 terms, the sum is still less than 20. Furthermore, to achieve a sum greater than 100, the number of terms required is 15092688622113788323693563264538101449859497

On a related point, the number of terms needed to equal or exceed $10^n$ are given in the sequence (OEIS A096618).

The first three terms are:
12367, 15092688622113788323693563264538101449859497 and $1.1×10^434$, .

share|cite|improve this answer

edited Jul 28 at 2:33

answered Jul 28 at 2:23

Martin Roberts

1,204318

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

The minimum number of terms needed so that harmonic series is at least $n$, can be found via oeis.org/A004080

The first few terms are:

1, 4, 11, 31, 83, 227, 616, 1674, 4550, 12367, 33617, 91380, 248397, 675214, 1835421, 4989191, 13562027, 36865412, 100210581, 272400600, 740461601, 2012783315, 5471312310, 14872568831, 40427833596, 109894245429, 298723530401, 812014744422

Thus,

The number of terms required for the sum to exceed 2 is 4.

The number of terms required for the sum to exceed 3 is 11.

The number of terms required for the sum to exceed 5 is 83.

Interestingly, after 2.5×10^8 terms, the sum is still less than 20. Furthermore, to achieve a sum greater than 100, the number of terms required is 15092688622113788323693563264538101449859497

On a related point, the number of terms needed to equal or exceed $10^n$ are given in the sequence (OEIS A096618).

The first three terms are:
12367, 15092688622113788323693563264538101449859497 and $1.1×10^434$, .

share|cite|improve this answer

edited Jul 28 at 2:33

answered Jul 28 at 2:23

Martin Roberts

1,204318

The minimum number of terms needed so that harmonic series is at least $n$, can be found via oeis.org/A004080

The first few terms are:

1, 4, 11, 31, 83, 227, 616, 1674, 4550, 12367, 33617, 91380, 248397, 675214, 1835421, 4989191, 13562027, 36865412, 100210581, 272400600, 740461601, 2012783315, 5471312310, 14872568831, 40427833596, 109894245429, 298723530401, 812014744422

Thus,

The number of terms required for the sum to exceed 2 is 4.

The number of terms required for the sum to exceed 3 is 11.

The number of terms required for the sum to exceed 5 is 83.

Interestingly, after 2.5×10^8 terms, the sum is still less than 20. Furthermore, to achieve a sum greater than 100, the number of terms required is 15092688622113788323693563264538101449859497

On a related point, the number of terms needed to equal or exceed $10^n$ are given in the sequence (OEIS A096618).

The first three terms are:
12367, 15092688622113788323693563264538101449859497 and $1.1×10^434$, .

share|cite|improve this answer

edited Jul 28 at 2:33

answered Jul 28 at 2:23

Martin Roberts

1,204318

share|cite|improve this answer

share|cite|improve this answer

edited Jul 28 at 2:33

edited Jul 28 at 2:33

edited Jul 28 at 2:33

answered Jul 28 at 2:23

Martin Roberts

1,204318

answered Jul 28 at 2:23

Martin Roberts

1,204318

answered Jul 28 at 2:23

Martin Roberts

1,204318

1,204318

add a comment | 

add a comment |