# Math Help - Who come to beat this ：）

1. Originally Posted by RockHard
Thanks for the confirmation, so this is best shown using the property of adding two series together will only converge if both series are convergent, correct?
That is NOT correct. For example, neither $\sum_{n=1}^\infty \frac{1}{n}$ nor $\sum_{n=1}^\infty -\frac{1}{n}$ converge but their sum is $\sum_{n=1}^\infty (\frac{1}{n}-\frac{1}{n})= \sum_{n=1}^\infty 0$ converges trivially.

If $\sum_{n=1}^\infty a_n$ diverges to positive infinity and $b_n> 0$, then $\sum_{n=1}^\infty (a_n+ b_n)$ diverges to infinity, by the comparison test, whether $\sum_{n=1}^\infty b_n$ converges or not.

In fact, I would have handled the initial problem by noting that, since $1+ \frac{1}{n}> 1$ $\left|(1+\frac{1}{n})a_n\right|> |a_n|$ and so $\sum_{n=1}^\infty (1+ \frac{1}{n})a_n$ diverges by the comparison test.

2. Originally Posted by HallsofIvy
since $1+ \frac{1}{n}> 1$ $\left|(1+\frac{1}{n})a_n\right|> |a_n|$ and so $\sum_{n=1}^\infty (1+ \frac{1}{n})a_n$ diverges by the comparison test.
I think you only proof $\sum_{n=1}^{\infty}\left|\left(1+\frac{1}{n}\right )a_{n}\right|$ diverges.

I think It's not imply $\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n}$
diverges

3. For example:

we take: $a_{n}=\left(-1\right)^n\frac{1}{n}$

then: $\sum_{n=1}^{\infty}\left|\left(1+\frac{1}{n}\right )a_{n}\right|=\sum_{n=1}^{\infty}\left|\left(1+\fr ac{1}{n}\right)(-1)^n\frac{1}{n}\right|\geq \sum_{n=1}^{\infty}\frac{1}{n}$

so it is diverges

but:
$\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n} =\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)(-1)^n\frac{1}{n}$

is a alternating series

so obviously converges

so $\sum_{n=1}^{\infty}\left|\left(1+\frac{1}{n}\right )a_{n}\right|$

diverges don't imply

$\sum_{n=1}^{\infty}\left(1+\frac{1}{n}\right)a_{n}$

diverges

4. The problem is equivalent to showing that if $\sum_n \left(1+\frac{1}{n}\right)a_n$ converges, then $\sum_n a_n$ converges.

Or, that if $\sum_n a_n$ converges, then $\sum_n\frac{a_n}{1+\frac{1}{n}}$ converges. (Applying this result to $\sum_n (1+\frac{1}{n})a_n$ gives the previous one)

So, assume that $\sum_n a_n$ converges.

Note that we have $\frac{1}{1+\frac{1}{n}}=1-\frac{1}{n}+O(\frac{1}{n^2})$, hence there are $n_0,C>0$ (that could be explicitated, but it's no use) such that, for all $n\geq n_0$,

$1-\frac{1}{n}-\frac{C}{n^2}\leq\frac{1}{1+\frac{1}{n}}\leq 1-\frac{1}{n}+\frac{C}{n^2}.$

However, since $\sum_n a_n$ converges, $a_n\to_n 0$ is bounded: $|a_n|\leq M$ for some $M$ and all $n$, hence $\sum_n \frac{a_n}{n^2}$ converges absolutely.

Moreover, it is a "classical" fact that $\sum_n \frac{a_n}{n}$ converges (cf. below).

As a consequence, we have

$\sum_{n=n_0}^N \frac{a_n}{1+\frac{1}{n}}-\sum_{n=n_0}^Na_n+\sum_{n=n_0}^N\frac{a_n}{n} = \sum_{n=n_0}^N C_n \frac{a_n}{n^2}$

where $|C_n|\leq C$. Each of the series converges, except (perhaps) the first one. Therefore, the first one converges as well.
.

This concludes.

Appendix: if $\sum_n a_n$ converges, then $\sum_n\frac{a_n}{n}$ converges. This is a consequence of "Abel summation". Let $S_n=a_1+\cdots+a_n$, so that $(S_n)_n$ converges to $\sum_{n=0}^\infty a_n$ and is therefore bounded: $|S_n|\leq M$ for all $n$ and some $M$. We have, for all $N$,

$\sum_{n=1}^N \frac{a_n}{n} = \sum_{n=1}^N \frac{S_n-S_{n-1}}{n} =$ $\sum_{n=1}^{N-1} \left(\frac{1}{n}-\frac{1}{n+1}\right)S_n + \frac{S_N}{N}=\sum_{n=1}^{N-1} \frac{1}{n(n+1)}S_n +\frac{S_N}{N}$,

and the term on the right end converges to a finite limit as $N\to\infty$ since $\frac{|S_n|}{n(n+1)}\leq \frac{M}{n^2}$ (general term of a convergent series), and $\frac{|S_N|}{N}\leq \frac{M}{N}\to_N 0$.

5. Excellent Work !!!

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