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The convergence of $sum_n=1^+infty bigg(tanbigg(fracpi2lambdabigg)bigg)^nbigg(n^2lambda+ sinbigg(frac(-1)^nnbigg)bigg)$
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Study the convergence of the following series for $|lambda|lt1, lambdainmathbbR$.
$$sum_n=1^+infty bigg(tanbigg(fracpi2lambdabigg)bigg)^nbigg(n^2lambda+ sinbigg(frac(-1)^nnbigg)bigg)$$
This series is giving me hard times, I have divided it into cases:
$$tanbigg(fracpi2lambdabigg) :
begincases
gt0, & textif 0 ltlambdalt1\[2ex]
lt0, & textif -1ltlambdalt0
endcases$$
$$2lambda :
begincases
gt1, & textif frac12ltlambdalt1 \[2ex]
gt0, lt1, & textif 0ltlambdaltfrac12 \[2ex]
lt0,gt-1 & textif -frac12ltlambdalt0 \[2ex]
lt-1, & textif -1ltlambdalt-frac12
endcases$$
besides we have to observe that $sinbigg(frac(-1)^nnbigg)$ converges for $nto+infty$
Does anyone have ideas and tips?
real-analysis sequences-and-series convergence
edited 2 days ago
Michael Hardy
204k23185460
asked 2 days ago
F.inc
1488
add a comment |Â
up vote
2
down vote
favorite
Study the convergence of the following series for $|lambda|lt1, lambdainmathbbR$.
$$sum_n=1^+infty bigg(tanbigg(fracpi2lambdabigg)bigg)^nbigg(n^2lambda+ sinbigg(frac(-1)^nnbigg)bigg)$$
This series is giving me hard times, I have divided it into cases:
$$tanbigg(fracpi2lambdabigg) :
begincases
gt0, & textif 0 ltlambdalt1\[2ex]
lt0, & textif -1ltlambdalt0
endcases$$
$$2lambda :
begincases
gt1, & textif frac12ltlambdalt1 \[2ex]
gt0, lt1, & textif 0ltlambdaltfrac12 \[2ex]
lt0,gt-1 & textif -frac12ltlambdalt0 \[2ex]
lt-1, & textif -1ltlambdalt-frac12
endcases$$
besides we have to observe that $sinbigg(frac(-1)^nnbigg)$ converges for $nto+infty$
Does anyone have ideas and tips?
real-analysis sequences-and-series convergence
edited 2 days ago
Michael Hardy
204k23185460
asked 2 days ago
F.inc
1488
See my edits to this question for proper MathJax usage.
– Michael Hardy
2 days ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Study the convergence of the following series for $|lambda|lt1, lambdainmathbbR$.
$$sum_n=1^+infty bigg(tanbigg(fracpi2lambdabigg)bigg)^nbigg(n^2lambda+ sinbigg(frac(-1)^nnbigg)bigg)$$
This series is giving me hard times, I have divided it into cases:
$$tanbigg(fracpi2lambdabigg) :
begincases
gt0, & textif 0 ltlambdalt1\[2ex]
lt0, & textif -1ltlambdalt0
endcases$$
$$2lambda :
begincases
gt1, & textif frac12ltlambdalt1 \[2ex]
gt0, lt1, & textif 0ltlambdaltfrac12 \[2ex]
lt0,gt-1 & textif -frac12ltlambdalt0 \[2ex]
lt-1, & textif -1ltlambdalt-frac12
endcases$$
besides we have to observe that $sinbigg(frac(-1)^nnbigg)$ converges for $nto+infty$
Does anyone have ideas and tips?
real-analysis sequences-and-series convergence
edited 2 days ago
Michael Hardy
204k23185460
asked 2 days ago
F.inc
1488
Study the convergence of the following series for $|lambda|lt1, lambdainmathbbR$.
$$sum_n=1^+infty bigg(tanbigg(fracpi2lambdabigg)bigg)^nbigg(n^2lambda+ sinbigg(frac(-1)^nnbigg)bigg)$$
This series is giving me hard times, I have divided it into cases:
$$tanbigg(fracpi2lambdabigg) :
begincases
gt0, & textif 0 ltlambdalt1\[2ex]
lt0, & textif -1ltlambdalt0
endcases$$
$$2lambda :
begincases
gt1, & textif frac12ltlambdalt1 \[2ex]
gt0, lt1, & textif 0ltlambdaltfrac12 \[2ex]
lt0,gt-1 & textif -frac12ltlambdalt0 \[2ex]
lt-1, & textif -1ltlambdalt-frac12
endcases$$
besides we have to observe that $sinbigg(frac(-1)^nnbigg)$ converges for $nto+infty$
Does anyone have ideas and tips?
real-analysis sequences-and-series convergence
edited 2 days ago
Michael Hardy
204k23185460
asked 2 days ago
F.inc
1488
edited 2 days ago
Michael Hardy
204k23185460
edited 2 days ago
Michael Hardy
204k23185460
edited 2 days ago
Michael Hardy
204k23185460
204k23185460
asked 2 days ago
F.inc
1488
asked 2 days ago
F.inc
1488
asked 2 days ago
F.inc
1488
1488
See my edits to this question for proper MathJax usage.
– Michael Hardy
2 days ago
add a comment |Â
See my edits to this question for proper MathJax usage.
– Michael Hardy
2 days ago
See my edits to this question for proper MathJax usage.
– Michael Hardy
2 days ago
See my edits to this question for proper MathJax usage.
– Michael Hardy
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Split the series into two parts.
For $sum tan^n (pi lambda /2) n^2lambda$:
$lambda > 1/2$: $tan(pi lambda /2)>1$, hence $$tan^n (pi lambda /2) n ^2lambda > tan^n (pi lambda /2) n to +infty [n to infty],$$ and the series diverges;
$lambda = 1/2$: $tan (pi lambda /2) n^2lambda = 1/n$, and $sum 1/n$ is the harmonic series, which diverges;
$0 < |lambda| < 1/2$: use the Cauchy root test:
$$
leftlvert tan^n left( frac pi lambda 2right) n^2lambdarightrvert^1/n = leftlvert tan left(frac pi lambda2right)rightrvert (n^1/n)^2lambda to leftlvert tan left(frac pilambda 2right) rightrvert < 1,
$$
hence the series converges;$lambda = -1/2$: this is a Leibniz series $sum (-1)^n n^-1$, hence converges;
$-1 < lambda < -1/2$: use Cauchy root test again, the root of absolute value tends to $-tan(pi lambda/2) > 1$, hence diverges.
Now for the second part:
Since $sin (1/n) sim 1/n$, or
$$
frac 2n pi leqslant sin left( frac 1n right) leqslant frac 1n
$$
for sufficiently large $n$ and the fact that $(1/n)^1/n to 1$, we have
$$
sin ^1/n left( frac 1 n right) to 1 quad [n to infty].
$$
Now by Cauchy root test, this series diverges when $|lambda| > 1/2$, converges when $|lambda| < 1/2$. When $lambda =1/2$, then the series is Leibniz type, hence converges; when $lambda = -1/2$, the series is
$sum sin (1/n)$. Since
$$
sin left( frac 1n right) geqslant frac 2 pi cdot frac 1 n,
$$ the series diverges.
In conclusion, the original series converges only when $|boldsymbollambda| < mathbf1/2$.
answered 2 days ago
xbh
9156
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Split the series into two parts.
For $sum tan^n (pi lambda /2) n^2lambda$:
$lambda > 1/2$: $tan(pi lambda /2)>1$, hence $$tan^n (pi lambda /2) n ^2lambda > tan^n (pi lambda /2) n to +infty [n to infty],$$ and the series diverges;
$lambda = 1/2$: $tan (pi lambda /2) n^2lambda = 1/n$, and $sum 1/n$ is the harmonic series, which diverges;
$0 < |lambda| < 1/2$: use the Cauchy root test:
$$
leftlvert tan^n left( frac pi lambda 2right) n^2lambdarightrvert^1/n = leftlvert tan left(frac pi lambda2right)rightrvert (n^1/n)^2lambda to leftlvert tan left(frac pilambda 2right) rightrvert < 1,
$$
hence the series converges;$lambda = -1/2$: this is a Leibniz series $sum (-1)^n n^-1$, hence converges;
$-1 < lambda < -1/2$: use Cauchy root test again, the root of absolute value tends to $-tan(pi lambda/2) > 1$, hence diverges.
Now for the second part:
Since $sin (1/n) sim 1/n$, or
$$
frac 2n pi leqslant sin left( frac 1n right) leqslant frac 1n
$$
for sufficiently large $n$ and the fact that $(1/n)^1/n to 1$, we have
$$
sin ^1/n left( frac 1 n right) to 1 quad [n to infty].
$$
Now by Cauchy root test, this series diverges when $|lambda| > 1/2$, converges when $|lambda| < 1/2$. When $lambda =1/2$, then the series is Leibniz type, hence converges; when $lambda = -1/2$, the series is
$sum sin (1/n)$. Since
$$
sin left( frac 1n right) geqslant frac 2 pi cdot frac 1 n,
$$ the series diverges.
In conclusion, the original series converges only when $|boldsymbollambda| < mathbf1/2$.
answered 2 days ago
xbh
9156
add a comment |Â
up vote
2
down vote
accepted
Split the series into two parts.
For $sum tan^n (pi lambda /2) n^2lambda$:
$lambda > 1/2$: $tan(pi lambda /2)>1$, hence $$tan^n (pi lambda /2) n ^2lambda > tan^n (pi lambda /2) n to +infty [n to infty],$$ and the series diverges;
$lambda = 1/2$: $tan (pi lambda /2) n^2lambda = 1/n$, and $sum 1/n$ is the harmonic series, which diverges;
$0 < |lambda| < 1/2$: use the Cauchy root test:
$$
leftlvert tan^n left( frac pi lambda 2right) n^2lambdarightrvert^1/n = leftlvert tan left(frac pi lambda2right)rightrvert (n^1/n)^2lambda to leftlvert tan left(frac pilambda 2right) rightrvert < 1,
$$
hence the series converges;$lambda = -1/2$: this is a Leibniz series $sum (-1)^n n^-1$, hence converges;
$-1 < lambda < -1/2$: use Cauchy root test again, the root of absolute value tends to $-tan(pi lambda/2) > 1$, hence diverges.
Now for the second part:
Since $sin (1/n) sim 1/n$, or
$$
frac 2n pi leqslant sin left( frac 1n right) leqslant frac 1n
$$
for sufficiently large $n$ and the fact that $(1/n)^1/n to 1$, we have
$$
sin ^1/n left( frac 1 n right) to 1 quad [n to infty].
$$
Now by Cauchy root test, this series diverges when $|lambda| > 1/2$, converges when $|lambda| < 1/2$. When $lambda =1/2$, then the series is Leibniz type, hence converges; when $lambda = -1/2$, the series is
$sum sin (1/n)$. Since
$$
sin left( frac 1n right) geqslant frac 2 pi cdot frac 1 n,
$$ the series diverges.
In conclusion, the original series converges only when $|boldsymbollambda| < mathbf1/2$.
answered 2 days ago
xbh
9156
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Split the series into two parts.
For $sum tan^n (pi lambda /2) n^2lambda$:
$lambda > 1/2$: $tan(pi lambda /2)>1$, hence $$tan^n (pi lambda /2) n ^2lambda > tan^n (pi lambda /2) n to +infty [n to infty],$$ and the series diverges;
$lambda = 1/2$: $tan (pi lambda /2) n^2lambda = 1/n$, and $sum 1/n$ is the harmonic series, which diverges;
$0 < |lambda| < 1/2$: use the Cauchy root test:
$$
leftlvert tan^n left( frac pi lambda 2right) n^2lambdarightrvert^1/n = leftlvert tan left(frac pi lambda2right)rightrvert (n^1/n)^2lambda to leftlvert tan left(frac pilambda 2right) rightrvert < 1,
$$
hence the series converges;$lambda = -1/2$: this is a Leibniz series $sum (-1)^n n^-1$, hence converges;
$-1 < lambda < -1/2$: use Cauchy root test again, the root of absolute value tends to $-tan(pi lambda/2) > 1$, hence diverges.
Now for the second part:
Since $sin (1/n) sim 1/n$, or
$$
frac 2n pi leqslant sin left( frac 1n right) leqslant frac 1n
$$
for sufficiently large $n$ and the fact that $(1/n)^1/n to 1$, we have
$$
sin ^1/n left( frac 1 n right) to 1 quad [n to infty].
$$
Now by Cauchy root test, this series diverges when $|lambda| > 1/2$, converges when $|lambda| < 1/2$. When $lambda =1/2$, then the series is Leibniz type, hence converges; when $lambda = -1/2$, the series is
$sum sin (1/n)$. Since
$$
sin left( frac 1n right) geqslant frac 2 pi cdot frac 1 n,
$$ the series diverges.
In conclusion, the original series converges only when $|boldsymbollambda| < mathbf1/2$.
answered 2 days ago
xbh
9156
Split the series into two parts.
For $sum tan^n (pi lambda /2) n^2lambda$:
$lambda > 1/2$: $tan(pi lambda /2)>1$, hence $$tan^n (pi lambda /2) n ^2lambda > tan^n (pi lambda /2) n to +infty [n to infty],$$ and the series diverges;
$lambda = 1/2$: $tan (pi lambda /2) n^2lambda = 1/n$, and $sum 1/n$ is the harmonic series, which diverges;
$0 < |lambda| < 1/2$: use the Cauchy root test:
$$
leftlvert tan^n left( frac pi lambda 2right) n^2lambdarightrvert^1/n = leftlvert tan left(frac pi lambda2right)rightrvert (n^1/n)^2lambda to leftlvert tan left(frac pilambda 2right) rightrvert < 1,
$$
hence the series converges;$lambda = -1/2$: this is a Leibniz series $sum (-1)^n n^-1$, hence converges;
$-1 < lambda < -1/2$: use Cauchy root test again, the root of absolute value tends to $-tan(pi lambda/2) > 1$, hence diverges.
Now for the second part:
Since $sin (1/n) sim 1/n$, or
$$
frac 2n pi leqslant sin left( frac 1n right) leqslant frac 1n
$$
for sufficiently large $n$ and the fact that $(1/n)^1/n to 1$, we have
$$
sin ^1/n left( frac 1 n right) to 1 quad [n to infty].
$$
Now by Cauchy root test, this series diverges when $|lambda| > 1/2$, converges when $|lambda| < 1/2$. When $lambda =1/2$, then the series is Leibniz type, hence converges; when $lambda = -1/2$, the series is
$sum sin (1/n)$. Since
$$
sin left( frac 1n right) geqslant frac 2 pi cdot frac 1 n,
$$ the series diverges.
In conclusion, the original series converges only when $|boldsymbollambda| < mathbf1/2$.
answered 2 days ago
xbh
9156
edited 2 days ago
answered 2 days ago
xbh
9156
answered 2 days ago
xbh
9156
answered 2 days ago
xbh
9156
9156
add a comment |Â
add a comment |Â
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See my edits to this question for proper MathJax usage.
– Michael Hardy
2 days ago