How do I prove that the series of $((1+\frac 1n)^n)a_n $ converges iff $\sum_n a_n$ converges?

Question

With sequences it simply arises from limit arithmetic, but with the series I cannot make it work. We have not learned integrals yet, so I can't use that. Any advice?

Answer 1

If you know the following result : If $a_n, b_n$ are real sequences such that the sums $S_n=\sum_{k=1}^n a_k$ are bounded and $b_n$ decrease to $0$, then the series $\sum a_nb_n$ is convergent, you can use that $b_n=e-(1+\frac{1}{n})^n$ is decreasing and $\to 0$, and if the series $a_n$ is convergent, then the $S_n$ are convergent and hence bounded. For the other implication, use $(1+\frac{1}{n})^{-n}-1/e$.

Answer 2

If the $a_n$ are positive, the result follows from the limit comparison test.

If the assumption is dropped, summation by parts can be used. Let us prove that if $\sum_n a_n$ converges, then $\sum_n \left(1+\frac 1n \right)^na_n$ converges. Let $A_n=\sum_{k=0}^n a_k$, so that $$\sum_{n=0}^N \left(1+\frac 1n \right)^na_n = A_N\left(1+\frac 1N \right)^N-\sum_{n=0}^{N-1}A_n\left( \left(1+\frac 1{n+1} \right)^{n+1}-\left(1+\frac 1n \right)^n\right)$$

Since $\lim_N \left(1+\frac 1N \right)^N = e$, $A_N\left(1+\frac 1N \right)^N$ converges. Since $A_N$ is bounded (because it converges) and $\left(1+\frac 1N \right)^N$ is increasing, $\sum_{n=0}^{N-1}A_n\left( \left(1+\frac 1{n+1} \right)^{n+1}-\left(1+\frac 1n \right)^n\right)$ is absolutely convergent, hence convergent.

$\sum_{n=0}^N \left(1+\frac 1n \right)^na_n$ is the sum of two convergent sequences, thus converges.

In the same fashion, one may prove that if $\sum_n a_n$ converges, then $\sum_n \left(1+\frac 1n \right)^{\color{red}{-n}}a_n$ converges. Apply this to $\sum_n \left(1+\frac 1n \right)^{n}a_n$ to get the converse of your statement.