Prove that if $\displaystyle \sum_{n=1}^ \infty na_n$ converges, then $\displaystyle\sum_{n=1}^ \infty a_n$ converges.
No, $a_n$ are not necessarily positive numbers.
I've been trying summation by parts.
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Think about what the summation converges to. If it converges to some real number then what would the sum of $n*a_n$ converge to? – Connor James Feb 03 '16 at 03:04
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I'd suggest writing $n=\sum_{k=1}^n 1$, then interchanging the order of summation on the partial sums. It should be clear what's happening to the terms to allow convergence. – Chappers Feb 03 '16 at 03:05
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@A.S. You can for the partial sums, since they're finite. – Chappers Feb 03 '16 at 03:09
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@Chap Then I get $\sum_{j=1}^N S_N-S_{j-1}$. I don't see how this is a simplification. – A.S. Feb 03 '16 at 03:51
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Let $S_n = \sum_{k=1}^n ka_k$ with $S_0 = 0$ and $T_n = \sum_{k=1}^n a_k$ then we have $$T_n = \sum_{k=1}^n \dfrac{S_k - S_{k-1}}{k} = \dfrac{S_n}{n} + \sum_{k=1}^{n-1}(\dfrac{1}{k} - \dfrac{1}{k+1})S_k = \dfrac{S_n}{n} + \sum_{k=1}^{n-1}\dfrac{1}{k(k+1)}S_k$$
Then it's clear that $T_n$ converges since $S_n$ converges and $\sum_{n=1}^\infty\dfrac{1}{n(n+1)}$ is finite
Petite Etincelle
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Collapsing a telescoped term to show convergence is unnecessary (imagine if we had a general monotonically increasing unbounded positive $f(n)$ instead of $n$). – A.S. Feb 03 '16 at 04:12
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@Jill_Johnson by checking coefficients before each $S_k$, since $S_k$ only appears in $\frac{S_{k+1} - S_k}{k+1}$ and $\frac{S_{k} - S_{k-1}}{k}$ – Petite Etincelle Feb 03 '16 at 09:42
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Dirichlet's Test with {$na_n$} and {$1/n$}?
- Since $\sum_{n=1}^\infty na_n$ converges, the partial sums $\sum_{n=1}^\infty na_n$ are bounded.
- $\lim \frac {1}{n} = 0$.
- The sequence $\frac {1}{n}$ is monotone, therefore the series $\sum_{n=1}^\infty \left\lvert \frac {1}{n+1} - \frac
{1}{n}\right\rvert$ converges and the conditions hold for the test.
So, $\sum_{n=1}^\infty (\frac {1}{n})(na_n)$=$\sum_{n=1}^\infty a_n$ converges.
Jill_Johnson
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Jill, the series for the third bullet point ought not have absolute values around the summands. - Mark – Mark Viola Feb 03 '16 at 03:39
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My textbook uses them. I think what I forgot was to place the absolute value signs are on the rest of the equation. I'll do that now. – Jill_Johnson Feb 03 '16 at 03:43
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Actually, supposedly 1/n works because it is monotone and bounded below by zero. – Jill_Johnson Feb 03 '16 at 03:46
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Jill, all one needs here is that $\left|\sum_{k=1}^{K}k a_k\right|\le B$ for some $B$ for all $K\ge 1$; not sure as to the purpose of the telescoping series. - Mark – Mark Viola Feb 03 '16 at 03:49