0
$\begingroup$

For what values of $x$ does $\sum_{n=0}^\infty n^7 x^n $ converge?

I have attempted to solve this problem but I am not sure of my solution - I'd be glad if you could review it and give me some sort of feedback.

Consider the series $$\sum_{n=0}^\infty |n^7x^n| = \sum_{n=0}^\infty|n^7||x|^n$$ This series converges absolutely for $-1<x<1$, and so the initial series converges for such $x$ as well.

Now, I check $-1$ and $1$
If $x=1$, then this series clearly diverges. When $x=-1$, then it does not converge (All of the tests that I know are inconclusive - my claim is based on my observations. Any hints how to prove this?)

Now it's time for $x < -1$ and $x>1$. In both cases. $n^7x^n$ does not approach zero, and so the series does not converge.

And so the series converges for $x \in (-1,1)$

$\endgroup$
1
$\begingroup$

At $-1$ again the terms do not go to zero so it cannot converge. Otherwise your solution looks right.

$\endgroup$
1
$\begingroup$

Depending on how you show absolute convergence, you won't need to check $x < -1$ and $x > 1$. You can apply the ratio test to show that the radius of convergence must be $1$. Since the center of this series is $0$, the interval of convergence is "at least" $(-1, 1)$ and "at most" $[-1,1]$. So now all you need to do is check $x = \pm 1$.

For $x = -1$, the $n$th term test for divergence is applicable. Recall that if $\displaystyle \lim_{n\to+\infty} a_n \ne 0$, then $\displaystyle \sum_{n=1}^{+\infty} a_n$ does not converge. Actually, this is what you used for the case $x < -1$ and $x > 1$.

$\endgroup$
  • $\begingroup$ In general the root test is more reliable than the ratio test, for example it easily gives you the correct behavior for $\sum_{n=0}^\infty x^{2^n}$. $\endgroup$ – Ian Nov 21 '17 at 19:41
0
$\begingroup$

One more test worth knowing.

The Alternating Series Test:

If all $a_n>0$ (or all $a_n <0$)

$S = \sum_\limits{n=0}^{\infty} (-1)^n a_n$ converges

If $a_n$ is monotonically decreasing (increasing if all $a_n$ are negative)

and $\lim_\limits{n \to \infty} a_n = 0$

Furthermore the partial sum

$S_k = \sum_\limits{n=0}^{k} (-1)^n a_n$ is approximates the (infinte) series with error bound $|S_k - S| < |a_{k+1}|$

For that matter, for any series $\sum a_n$ to converge

$\lim_\limits{n\to \infty} a_n = 0$ is a necessary condition.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.