Prove $f_n=sum_{k=0}^{infty}a_k(1-frac{1}{n})^k$ converges to $infty$.
$begingroup$
Given that $;sum a_k$ is a divergent series in $(0,infty)$ and $sum a_kX^k$ has radius of convergence $rho_a=1$.
This is an exercise from Amann Herbert Analysis.
It gives a hint that we can use the Bernoulli inequality to get an upper bound for terms of the form $1-(1-frac{1}{n})^k$.
sequences-and-series analysis power-series
asked Jan 18 at 3:48
Tao XTao X
575
$endgroup$
add a comment |
$begingroup$
Given that $;sum a_k$ is a divergent series in $(0,infty)$ and $sum a_kX^k$ has radius of convergence $rho_a=1$.
This is an exercise from Amann Herbert Analysis.
It gives a hint that we can use the Bernoulli inequality to get an upper bound for terms of the form $1-(1-frac{1}{n})^k$.
sequences-and-series analysis power-series
asked Jan 18 at 3:48
Tao XTao X
575
$endgroup$
1
$begingroup$
Start by reading How to ask a good question. Some personal input is necessary. Can you make an attempt at the application of Bernoull's inequality to that expression and add it to your post?
$endgroup$
– RRL
Jan 18 at 5:02
$begingroup$
Thanks. I have read it before, but I didn't have any idea about this problem.
$endgroup$
– Tao X
Jan 18 at 13:00
add a comment |
$begingroup$
Given that $;sum a_k$ is a divergent series in $(0,infty)$ and $sum a_kX^k$ has radius of convergence $rho_a=1$.
This is an exercise from Amann Herbert Analysis.
It gives a hint that we can use the Bernoulli inequality to get an upper bound for terms of the form $1-(1-frac{1}{n})^k$.
sequences-and-series analysis power-series
asked Jan 18 at 3:48
Tao XTao X
575
$endgroup$
Given that $;sum a_k$ is a divergent series in $(0,infty)$ and $sum a_kX^k$ has radius of convergence $rho_a=1$.
This is an exercise from Amann Herbert Analysis.
It gives a hint that we can use the Bernoulli inequality to get an upper bound for terms of the form $1-(1-frac{1}{n})^k$.
sequences-and-series analysis power-series
sequences-and-series analysis power-series
asked Jan 18 at 3:48
Tao XTao X
575
asked Jan 18 at 3:48
Tao XTao X
575
asked Jan 18 at 3:48
Tao XTao X
575
asked Jan 18 at 3:48
Tao XTao X
575
asked Jan 18 at 3:48
Tao XTao X
575
575
1
$begingroup$
Start by reading How to ask a good question. Some personal input is necessary. Can you make an attempt at the application of Bernoull's inequality to that expression and add it to your post?
$endgroup$
– RRL
Jan 18 at 5:02
$begingroup$
Thanks. I have read it before, but I didn't have any idea about this problem.
$endgroup$
– Tao X
Jan 18 at 13:00
add a comment |
1
$begingroup$
Start by reading How to ask a good question. Some personal input is necessary. Can you make an attempt at the application of Bernoull's inequality to that expression and add it to your post?
$endgroup$
– RRL
Jan 18 at 5:02
$begingroup$
Thanks. I have read it before, but I didn't have any idea about this problem.
$endgroup$
– Tao X
Jan 18 at 13:00
1
1
$begingroup$
Start by reading How to ask a good question. Some personal input is necessary. Can you make an attempt at the application of Bernoull's inequality to that expression and add it to your post?
$endgroup$
– RRL
Jan 18 at 5:02
$begingroup$
Start by reading How to ask a good question. Some personal input is necessary. Can you make an attempt at the application of Bernoull's inequality to that expression and add it to your post?
$endgroup$
– RRL
Jan 18 at 5:02
$begingroup$
Thanks. I have read it before, but I didn't have any idea about this problem.
$endgroup$
– Tao X
Jan 18 at 13:00
$begingroup$
Thanks. I have read it before, but I didn't have any idea about this problem.
$endgroup$
– Tao X
Jan 18 at 13:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $M in (0,infty)$. Choose $N$ such that $ sumlimits_{k=1}^{N} a_k >2M$. Observe that $ sumlimits_{k=1}^{N} a_k [1-(1-frac 1 n)^{k}] <frac 1 2sumlimits_{k=1}^{N} a_k $ for $n$ sufficiently large. Hence $ sumlimits_{k=1}^{N} a_k (1-frac 1 n)^{k}>frac 1 2sumlimits_{k=1}^{N} a_k >M$. It follows that $ sumlimits_{k=1}^{infty} a_k (1-frac 1 n)^{k}>M$ for $n$ sufficiently large.
answered Jan 18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
Excellent,thanks. It seems that I didn't understand what $sum a_k$ is divergent in $(0,infty)$ means.
$endgroup$
– Tao X
Jan 18 at 12:59
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
Let $M in (0,infty)$. Choose $N$ such that $ sumlimits_{k=1}^{N} a_k >2M$. Observe that $ sumlimits_{k=1}^{N} a_k [1-(1-frac 1 n)^{k}] <frac 1 2sumlimits_{k=1}^{N} a_k $ for $n$ sufficiently large. Hence $ sumlimits_{k=1}^{N} a_k (1-frac 1 n)^{k}>frac 1 2sumlimits_{k=1}^{N} a_k >M$. It follows that $ sumlimits_{k=1}^{infty} a_k (1-frac 1 n)^{k}>M$ for $n$ sufficiently large.
answered Jan 18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
Excellent,thanks. It seems that I didn't understand what $sum a_k$ is divergent in $(0,infty)$ means.
$endgroup$
– Tao X
Jan 18 at 12:59
add a comment |
$begingroup$
Let $M in (0,infty)$. Choose $N$ such that $ sumlimits_{k=1}^{N} a_k >2M$. Observe that $ sumlimits_{k=1}^{N} a_k [1-(1-frac 1 n)^{k}] <frac 1 2sumlimits_{k=1}^{N} a_k $ for $n$ sufficiently large. Hence $ sumlimits_{k=1}^{N} a_k (1-frac 1 n)^{k}>frac 1 2sumlimits_{k=1}^{N} a_k >M$. It follows that $ sumlimits_{k=1}^{infty} a_k (1-frac 1 n)^{k}>M$ for $n$ sufficiently large.
answered Jan 18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
Excellent,thanks. It seems that I didn't understand what $sum a_k$ is divergent in $(0,infty)$ means.
$endgroup$
– Tao X
Jan 18 at 12:59
add a comment |
$begingroup$
Let $M in (0,infty)$. Choose $N$ such that $ sumlimits_{k=1}^{N} a_k >2M$. Observe that $ sumlimits_{k=1}^{N} a_k [1-(1-frac 1 n)^{k}] <frac 1 2sumlimits_{k=1}^{N} a_k $ for $n$ sufficiently large. Hence $ sumlimits_{k=1}^{N} a_k (1-frac 1 n)^{k}>frac 1 2sumlimits_{k=1}^{N} a_k >M$. It follows that $ sumlimits_{k=1}^{infty} a_k (1-frac 1 n)^{k}>M$ for $n$ sufficiently large.
answered Jan 18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
Let $M in (0,infty)$. Choose $N$ such that $ sumlimits_{k=1}^{N} a_k >2M$. Observe that $ sumlimits_{k=1}^{N} a_k [1-(1-frac 1 n)^{k}] <frac 1 2sumlimits_{k=1}^{N} a_k $ for $n$ sufficiently large. Hence $ sumlimits_{k=1}^{N} a_k (1-frac 1 n)^{k}>frac 1 2sumlimits_{k=1}^{N} a_k >M$. It follows that $ sumlimits_{k=1}^{infty} a_k (1-frac 1 n)^{k}>M$ for $n$ sufficiently large.
answered Jan 18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
answered Jan 18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
answered Jan 18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
answered Jan 18 at 6:09
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
59.5k42161
$begingroup$
Excellent,thanks. It seems that I didn't understand what $sum a_k$ is divergent in $(0,infty)$ means.
$endgroup$
– Tao X
Jan 18 at 12:59
add a comment |
$begingroup$
Excellent,thanks. It seems that I didn't understand what $sum a_k$ is divergent in $(0,infty)$ means.
$endgroup$
– Tao X
Jan 18 at 12:59
$begingroup$
Excellent,thanks. It seems that I didn't understand what $sum a_k$ is divergent in $(0,infty)$ means.
$endgroup$
– Tao X
Jan 18 at 12:59
$begingroup$
Excellent,thanks. It seems that I didn't understand what $sum a_k$ is divergent in $(0,infty)$ means.
$endgroup$
– Tao X
Jan 18 at 12:59
add a comment |
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$begingroup$
Start by reading How to ask a good question. Some personal input is necessary. Can you make an attempt at the application of Bernoull's inequality to that expression and add it to your post?
$endgroup$
– RRL
Jan 18 at 5:02
$begingroup$
Thanks. I have read it before, but I didn't have any idea about this problem.
$endgroup$
– Tao X
Jan 18 at 13:00