Compute $sum_ngeq 1u_n$ where $u_n=frac1n$ if $nnotequiv 0pmod 3$ and $frac-2n$ if $nequiv 0pmod 3$.

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Compute $sum_ngeq 1u_n$ where $$u_n=begincases
frac1n&nnotequiv 0pmod 3\
frac-2n&nequiv 0pmod 3endcases.$$

I really tried many thing, but not concludive. I set $$v_p=frac13p-2+frac13p-2-frac23p.$$
Therefore
$$sum_kgeq 1^3nu_k=sum_k=1^n v_k=sum_k=1^nfrac13k+sum_k=1^nfrac13k-1+sum_k=1^nfrac13k-2.$$
Then I tried to get the same denominator, but I gut stuck because the $3k-1$ and $3k-2$.

sequences-and-series

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asked Jul 22 at 12:59

MathBeginner

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Compute $sum_ngeq 1u_n$ where $$u_n=begincases
frac1n&nnotequiv 0pmod 3\
frac-2n&nequiv 0pmod 3endcases.$$

I really tried many thing, but not concludive. I set $$v_p=frac13p-2+frac13p-2-frac23p.$$
Therefore
$$sum_kgeq 1^3nu_k=sum_k=1^n v_k=sum_k=1^nfrac13k+sum_k=1^nfrac13k-1+sum_k=1^nfrac13k-2.$$
Then I tried to get the same denominator, but I gut stuck because the $3k-1$ and $3k-2$.

sequences-and-series

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asked Jul 22 at 12:59

MathBeginner

707312

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up vote
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favorite

up vote
0
down vote

favorite

Compute $sum_ngeq 1u_n$ where $$u_n=begincases
frac1n&nnotequiv 0pmod 3\
frac-2n&nequiv 0pmod 3endcases.$$

I really tried many thing, but not concludive. I set $$v_p=frac13p-2+frac13p-2-frac23p.$$
Therefore
$$sum_kgeq 1^3nu_k=sum_k=1^n v_k=sum_k=1^nfrac13k+sum_k=1^nfrac13k-1+sum_k=1^nfrac13k-2.$$
Then I tried to get the same denominator, but I gut stuck because the $3k-1$ and $3k-2$.

sequences-and-series

share|cite|improve this question

asked Jul 22 at 12:59

MathBeginner

707312

Compute $sum_ngeq 1u_n$ where $$u_n=begincases
frac1n&nnotequiv 0pmod 3\
frac-2n&nequiv 0pmod 3endcases.$$

I really tried many thing, but not concludive. I set $$v_p=frac13p-2+frac13p-2-frac23p.$$
Therefore
$$sum_kgeq 1^3nu_k=sum_k=1^n v_k=sum_k=1^nfrac13k+sum_k=1^nfrac13k-1+sum_k=1^nfrac13k-2.$$
Then I tried to get the same denominator, but I gut stuck because the $3k-1$ and $3k-2$.

sequences-and-series

share|cite|improve this question

asked Jul 22 at 12:59

MathBeginner

707312

share|cite|improve this question

share|cite|improve this question

asked Jul 22 at 12:59

MathBeginner

707312

asked Jul 22 at 12:59

MathBeginner

707312

asked Jul 22 at 12:59

MathBeginner

707312

707312

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2 Answers
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You mean
$$sum_k=1^3nu_k=sum_k=1^nfrac13k-2
+sum_k=1^nfrac13k-1-2sum_k=1^nfrac13k
=H_3n-H_n$$
where $H_n=sum_k=1^n1/k$. Then $H_n=ln n+gamma+o(1)$ and so
$$sum_k=1^3nu_k=ln 3+o(1)$$
etc.

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answered Jul 22 at 13:04

Lord Shark the Unknown

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Hint

You also have $$v_p=frac13k-2+frac13k-1+frac13k-frac33k,$$
i.e. $$sum_k=1^n v_k=sum_k=1^3nfrac1k-sum_k=1^nfrac1k.$$

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answered Jul 22 at 13:02

Surb

36.3k84274

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

You mean
$$sum_k=1^3nu_k=sum_k=1^nfrac13k-2
+sum_k=1^nfrac13k-1-2sum_k=1^nfrac13k
=H_3n-H_n$$
where $H_n=sum_k=1^n1/k$. Then $H_n=ln n+gamma+o(1)$ and so
$$sum_k=1^3nu_k=ln 3+o(1)$$
etc.

share|cite|improve this answer

answered Jul 22 at 13:04

Lord Shark the Unknown

85.2k950111

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up vote
1
down vote

You mean
$$sum_k=1^3nu_k=sum_k=1^nfrac13k-2
+sum_k=1^nfrac13k-1-2sum_k=1^nfrac13k
=H_3n-H_n$$
where $H_n=sum_k=1^n1/k$. Then $H_n=ln n+gamma+o(1)$ and so
$$sum_k=1^3nu_k=ln 3+o(1)$$
etc.

share|cite|improve this answer

answered Jul 22 at 13:04

Lord Shark the Unknown

85.2k950111

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up vote
1
down vote

up vote
1
down vote

You mean
$$sum_k=1^3nu_k=sum_k=1^nfrac13k-2
+sum_k=1^nfrac13k-1-2sum_k=1^nfrac13k
=H_3n-H_n$$
where $H_n=sum_k=1^n1/k$. Then $H_n=ln n+gamma+o(1)$ and so
$$sum_k=1^3nu_k=ln 3+o(1)$$
etc.

share|cite|improve this answer

answered Jul 22 at 13:04

Lord Shark the Unknown

85.2k950111

You mean
$$sum_k=1^3nu_k=sum_k=1^nfrac13k-2
+sum_k=1^nfrac13k-1-2sum_k=1^nfrac13k
=H_3n-H_n$$
where $H_n=sum_k=1^n1/k$. Then $H_n=ln n+gamma+o(1)$ and so
$$sum_k=1^3nu_k=ln 3+o(1)$$
etc.

share|cite|improve this answer

answered Jul 22 at 13:04

Lord Shark the Unknown

85.2k950111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 13:04

Lord Shark the Unknown

85.2k950111

answered Jul 22 at 13:04

Lord Shark the Unknown

85.2k950111

answered Jul 22 at 13:04

Lord Shark the Unknown

85.2k950111

85.2k950111

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add a comment | 

up vote
0
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Hint

You also have $$v_p=frac13k-2+frac13k-1+frac13k-frac33k,$$
i.e. $$sum_k=1^n v_k=sum_k=1^3nfrac1k-sum_k=1^nfrac1k.$$

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answered Jul 22 at 13:02

Surb

36.3k84274

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up vote
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Hint

You also have $$v_p=frac13k-2+frac13k-1+frac13k-frac33k,$$
i.e. $$sum_k=1^n v_k=sum_k=1^3nfrac1k-sum_k=1^nfrac1k.$$

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answered Jul 22 at 13:02

Surb

36.3k84274

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up vote
0
down vote

up vote
0
down vote

Hint

You also have $$v_p=frac13k-2+frac13k-1+frac13k-frac33k,$$
i.e. $$sum_k=1^n v_k=sum_k=1^3nfrac1k-sum_k=1^nfrac1k.$$

share|cite|improve this answer

answered Jul 22 at 13:02

Surb

36.3k84274

Hint

You also have $$v_p=frac13k-2+frac13k-1+frac13k-frac33k,$$
i.e. $$sum_k=1^n v_k=sum_k=1^3nfrac1k-sum_k=1^nfrac1k.$$

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answered Jul 22 at 13:02

Surb

36.3k84274

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share|cite|improve this answer

answered Jul 22 at 13:02

Surb

36.3k84274

answered Jul 22 at 13:02

Surb

36.3k84274

answered Jul 22 at 13:02

Surb

36.3k84274

36.3k84274

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