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Suppose that the series of positive numbers $\sum a_n$ can be shown to converge using either the Ratio Test or the Root Test. Show that the series $\sum n^k a_n$ converges for any positive integer $k$.

Is my proof sufficient?

Proof:

Assume $\sum a_n$ converges by the Ratio Test.

That is $R<1$ for $\varlimsup_{n\to\infty} \frac{a_{n+1}}{a_n}$.

Consider $\sum n^k a_n$, $k\in \mathbb{Z^+}$.

We have $\varlimsup_{n\to\infty} \frac{(n+1)^k a_{n+1}}{n^k a_n}$.

By Induction, $\sum \frac{(n+1)^k a_{n+1}}{n^k a_n} \leq$ $\max\{(n+1)^k\}$ $\frac{a_{n+1}}{a_n} < 1$ , for sufficiently large $k$.

Hence by Ratio Test, $\sum n^k a_n$ converges.

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    $\begingroup$ The idea is right, we want to show that $\limsup \frac{(n+1)^k|a_{n+1}|}{n^k|a_n|} \lt 1$, but that has not been done. $\endgroup$ – André Nicolas Apr 5 '16 at 4:02
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    $\begingroup$ With $a_n = {1 \over n^2}$ the series converges but diverges for $k=2$. $\endgroup$ – copper.hat Apr 5 '16 at 4:02
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    $\begingroup$ @copper.hat: The hypothesis is that the convergence of $\sum a_n$ can be proved using Ratio or Root Test. These are inconclusive for $a_n=\frac{1}{n^2}$. $\endgroup$ – André Nicolas Apr 5 '16 at 4:04
  • $\begingroup$ @AndréNicolas: Thanks, I was rather puzzled. I need to read more carefully. $\endgroup$ – copper.hat Apr 5 '16 at 4:04
  • $\begingroup$ I know many convergence tests but not all of them by name. What is the Root Test? $\endgroup$ – DanielWainfleet Apr 5 '16 at 4:22
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Hint:

T. Bongers makes good points. Here is a hint to salvage your proof. $$\lim_{n\to\infty}\frac{(n+1)^k |a_{n+1}|}{n^k |a_k|}=\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^k\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.$$

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As it stands, this is not sufficient. For example, you have the line

That is $R<1$ for $\varlimsup_{n\to\infty} \frac{a_{n+1}}{a_n}$.

What is $R$, and what does it have to do with the terms? I assume that you want to define $R$ to be this limit supremum.

We have $\varlimsup_{n\to\infty} \frac{(n+1)^k a_{n+1}}{n^k a_n}$.

I assume that you're aiming to have a bound on this quantity, but you don't. The bound here ought to be $R$, and that needs to be proven - you'll have to study the quantity $$\frac{(n + 1)^k}{n^k}$$ carefully to conclude this.

Then you have

By Induction, $\sum \frac{(n+1)^k a_{n+1}}{n^k a_n} \leq$ $\max\{(n+1)^k\}$ $\frac{a_{n+1}}{a_n} < 1$ , for sufficiently large $k$.

This doesn't really make any sense. If $k$ gets large, the quantity in the maximum gets larger, not smaller, and your bound doesn't work. Moreover, you're using $n$ as the variable of summation and then as something fixed, so I'm really not even sure how to parse this. It's also not clear what induction has to do with any of this.


As a very brief sketch of how to fix this: If you can show that

$$\sum_{n = 1}^{\infty} n^k R^n$$

converges for any $|R| < 1$, you're pretty much done - then some sort of comparison test will finish it. This can be done with the ratio test.

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    $\begingroup$ In the Q the phrase "for sufficiently large k" should be "for sufficiently large n". This may have been a typo bt the OP. $\endgroup$ – DanielWainfleet Apr 5 '16 at 4:21
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First note that (as @T.Bongers pointed out) if $|R|<1$ and $k$ is a positive integer, then $$\lim_{n\to\infty} |n^kR^n|^{\frac1n}=R\lim_{n\to\infty}n^{\frac kn}=R\left(\lim_{n\to\infty}n^{\frac1n} \right)^k=R<1,$$ so by the root test, $\sum_n n^k R^n$ converges absolutely.

Now, suppose $\{a_n\}$ is a sequence of real numbers such that $$\limsup_{n\to\infty} |a_n|^{\frac1n}=\rho\in(0,1). $$ Let $R\in(\rho,1)$. Choose $N$ so that $n\geqslant N$ implies $|a_n|^{\frac1n}<R$, and hence $|a_n|<R^n$. It follows that $n^k|a_n|\leqslant n^kR^n$ for $n\geqslant N$, and so by basic comparison, $\sum_n n^k a_n$ converges absolutely.

Since $$\limsup_{n\to\infty} c_n^{\frac1n}\leqslant\limsup_{n\to\infty}\frac{c_{n+1}}{c_n} $$ for any positive sequence $\{c_n\}$, whenever the ratio test shows convergence, the root test does too, and so we need only consider the root test here.

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