How to prove that $\sum_{n=1}^{\infty} n^{\frac{1}{n}} a_n$ is convergent. Where $\sum_{n=1}^{\infty}a_n$ is convergent.
I know that
Abel's Test: Let $\sum_{n=1}^{\infty}a_n$ is convergent and the sequence $<b_n>$ is monotonic and bounded then $\sum_{n=1}^{\infty}a_nb_n$ is convergent. I don't know how to apply this test. Please help me.
Consider $\sum (n^{1/n}-1)a_n$. Then $\lim_{n\to\infty} n^{1/n}-1=0$ and $n^{1/n}-1$ is eventually monotonically decreasing (for $n\ge 4$), moreover the partial sums of $\sum a_n$ are bounded. So by Abel-Dirichlet test $\sum (n^{1/n}-1)a_n$ is converging. Adding $\sum a_n$ we get that $\sum n^{1/n}a_n$ is converging.
Proof that $n^{1/n}$ is decreasing for $n\ge 4$.
$$n^{1/n}\quad ? \quad (n+1)^{1/(n+1)}$$ iff $$n^{n+1}\quad ? \quad (n+1)^n$$ where $?$ is $"<"$ or $">"$ (we shall prove that $?=">"$).
By the binomial formula,
\begin{align*} (n+1)^n &= n^n+nn^{n-1}+\frac{n(n-1)}{2} n^{n-2}+\dots+\frac{n(n-1)\cdots(n-k+1)}{k!}n^{n-k}+\dots+1 \\ &< n^n+\frac{n^n}{1!}+\frac{n^n}{2!}+\cdots+\frac{n^n}{k!}+\cdots+\frac{n^n}{n!} \\ &\le n^n(1+1+\frac1{2!}+\cdots+\frac1{n!}) \\ &< n^n(1+e). \end{align*} So $n^{1/n}>(1+n)^{1/(1+n)}$ if $n>1+e$ and in fact $?=">"$ indeed.
Hint:
$$ n^{1/n} \ge 1 \iff n \ge 1^n$$
and
$$ n^{1/n} \le 2 \iff n \le 2^n$$
In other words $n^{1/n} \in [1,2]$ for $n \ge 1$.
A good place to start understanding the function $x^{1/x}$ is to graph it. You can see visually that $x^{1/x} \in [1,2]$ for $x \ge 1$ and you can see that it is decreasing if $x \ge e$ (2.718...). You can then try to prove these statements analytically (e.g. compute the derivative).
