Mixing
Application of Leibniz Criterion to $sum_{k=2}^{+infty}(k^2+1)sin(frac{1}{k^3-1})alpha^k$
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I have to discuss absolute and simple convergence of $sum_{k=2}^{+infty}(k^2+1)sin(frac{1}{k^3-1})alpha^k$,where $alphainmathbb{R}$.
If $alpha=-1$ I have $sum_{k=2}^{+infty}(-1)^kb_k$,where $b_k=(k^2+1)sin(frac{1}{k^3-1})$.By Leibniz Criterion,for convergence I must have:
i)$lim_{kto+infty}b_k=0$
ii)$b_k$ eventually decreasing for $kto+infty$
For the first condition $lim_{kto+infty}=(k^2+1)sin(frac{1}{k^3-1})simlim_{kto+infty}frac{1}{k}=0$
As for the second condition ${frac{1}{k}}'=-frac{1}{k^2}lt0$.Can I deduce from this that our original $b_k$ is also decreasing,or I have to study derivative of $b_k$ itself?
real-analysis calculus sequences-and-series convergence
asked Jan 29 at 0:23
Turan NasibliTuran Nasibli
846
$endgroup$
add a comment |
$begingroup$
I have to discuss absolute and simple convergence of $sum_{k=2}^{+infty}(k^2+1)sin(frac{1}{k^3-1})alpha^k$,where $alphainmathbb{R}$.
If $alpha=-1$ I have $sum_{k=2}^{+infty}(-1)^kb_k$,where $b_k=(k^2+1)sin(frac{1}{k^3-1})$.By Leibniz Criterion,for convergence I must have:
i)$lim_{kto+infty}b_k=0$
ii)$b_k$ eventually decreasing for $kto+infty$
For the first condition $lim_{kto+infty}=(k^2+1)sin(frac{1}{k^3-1})simlim_{kto+infty}frac{1}{k}=0$
As for the second condition ${frac{1}{k}}'=-frac{1}{k^2}lt0$.Can I deduce from this that our original $b_k$ is also decreasing,or I have to study derivative of $b_k$ itself?
real-analysis calculus sequences-and-series convergence
asked Jan 29 at 0:23
Turan NasibliTuran Nasibli
846
$endgroup$
add a comment |
$begingroup$
I have to discuss absolute and simple convergence of $sum_{k=2}^{+infty}(k^2+1)sin(frac{1}{k^3-1})alpha^k$,where $alphainmathbb{R}$.
If $alpha=-1$ I have $sum_{k=2}^{+infty}(-1)^kb_k$,where $b_k=(k^2+1)sin(frac{1}{k^3-1})$.By Leibniz Criterion,for convergence I must have:
i)$lim_{kto+infty}b_k=0$
ii)$b_k$ eventually decreasing for $kto+infty$
For the first condition $lim_{kto+infty}=(k^2+1)sin(frac{1}{k^3-1})simlim_{kto+infty}frac{1}{k}=0$
As for the second condition ${frac{1}{k}}'=-frac{1}{k^2}lt0$.Can I deduce from this that our original $b_k$ is also decreasing,or I have to study derivative of $b_k$ itself?
real-analysis calculus sequences-and-series convergence
asked Jan 29 at 0:23
Turan NasibliTuran Nasibli
846
$endgroup$
I have to discuss absolute and simple convergence of $sum_{k=2}^{+infty}(k^2+1)sin(frac{1}{k^3-1})alpha^k$,where $alphainmathbb{R}$.
If $alpha=-1$ I have $sum_{k=2}^{+infty}(-1)^kb_k$,where $b_k=(k^2+1)sin(frac{1}{k^3-1})$.By Leibniz Criterion,for convergence I must have:
i)$lim_{kto+infty}b_k=0$
ii)$b_k$ eventually decreasing for $kto+infty$
For the first condition $lim_{kto+infty}=(k^2+1)sin(frac{1}{k^3-1})simlim_{kto+infty}frac{1}{k}=0$
As for the second condition ${frac{1}{k}}'=-frac{1}{k^2}lt0$.Can I deduce from this that our original $b_k$ is also decreasing,or I have to study derivative of $b_k$ itself?
real-analysis calculus sequences-and-series convergence
real-analysis calculus sequences-and-series convergence
asked Jan 29 at 0:23
Turan NasibliTuran Nasibli
846
asked Jan 29 at 0:23
Turan NasibliTuran Nasibli
846
edited Jan 29 at 1:18
Turan Nasibli
asked Jan 29 at 0:23
Turan NasibliTuran Nasibli
846
asked Jan 29 at 0:23
Turan NasibliTuran Nasibli
846
asked Jan 29 at 0:23
Turan NasibliTuran Nasibli
846
846
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1 Answer
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By applying ratio test, we know that the series is absolutely convergent for $|alpha|<1$ and divergent for $|alpha|>1$. For $alpha=1 $ the series is divergent since $$(k^2+1)sin(frac{1}{k^3-1})sim {k^2+1over k^3-1}sim {1over k}$$similarly for $alpha=-1$$$(k^2+1)sin(frac{1}{k^3-1})(-1)^ksim {(-1)^kover k}$$which is convergent but not absolutely.
answered Feb 2 at 16:00
Mostafa AyazMostafa Ayaz
18.2k31040
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1 Answer
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1 Answer
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$begingroup$
By applying ratio test, we know that the series is absolutely convergent for $|alpha|<1$ and divergent for $|alpha|>1$. For $alpha=1 $ the series is divergent since $$(k^2+1)sin(frac{1}{k^3-1})sim {k^2+1over k^3-1}sim {1over k}$$similarly for $alpha=-1$$$(k^2+1)sin(frac{1}{k^3-1})(-1)^ksim {(-1)^kover k}$$which is convergent but not absolutely.
answered Feb 2 at 16:00
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
$begingroup$
By applying ratio test, we know that the series is absolutely convergent for $|alpha|<1$ and divergent for $|alpha|>1$. For $alpha=1 $ the series is divergent since $$(k^2+1)sin(frac{1}{k^3-1})sim {k^2+1over k^3-1}sim {1over k}$$similarly for $alpha=-1$$$(k^2+1)sin(frac{1}{k^3-1})(-1)^ksim {(-1)^kover k}$$which is convergent but not absolutely.
answered Feb 2 at 16:00
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
add a comment |
$begingroup$
By applying ratio test, we know that the series is absolutely convergent for $|alpha|<1$ and divergent for $|alpha|>1$. For $alpha=1 $ the series is divergent since $$(k^2+1)sin(frac{1}{k^3-1})sim {k^2+1over k^3-1}sim {1over k}$$similarly for $alpha=-1$$$(k^2+1)sin(frac{1}{k^3-1})(-1)^ksim {(-1)^kover k}$$which is convergent but not absolutely.
answered Feb 2 at 16:00
Mostafa AyazMostafa Ayaz
18.2k31040
$endgroup$
By applying ratio test, we know that the series is absolutely convergent for $|alpha|<1$ and divergent for $|alpha|>1$. For $alpha=1 $ the series is divergent since $$(k^2+1)sin(frac{1}{k^3-1})sim {k^2+1over k^3-1}sim {1over k}$$similarly for $alpha=-1$$$(k^2+1)sin(frac{1}{k^3-1})(-1)^ksim {(-1)^kover k}$$which is convergent but not absolutely.
answered Feb 2 at 16:00
Mostafa AyazMostafa Ayaz
18.2k31040
answered Feb 2 at 16:00
Mostafa AyazMostafa Ayaz
18.2k31040
answered Feb 2 at 16:00
Mostafa AyazMostafa Ayaz
18.2k31040
answered Feb 2 at 16:00
Mostafa AyazMostafa Ayaz
18.2k31040
18.2k31040
add a comment |
add a comment |
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