How find the $\frac{1}{e}\le R\le 1$

Question

Qustion:

let $a_{n}> 0$,and such $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converge,let $$b_{m}=\sum_{n=1}^{\infty}\left(1+\dfrac{1}{n^m}\right)^na_{n}$$.

show that $$\dfrac{1}{e}\le R\le 1$$ where the $R$ is the radius of convergence of $\displaystyle\sum_{m=1}^{\infty}b_{m}x^m$

My try:I know this therom:

If the radius of convergence of $\sum_{n=0}^{\infty}a_{n}x^n$ and $\sum_{n=0}^{\infty}b_{n}x^n$ are $R_{1}$ and $R_{2}$, then

the radius of convergence $R$ of $\sum_{n=0}^{\infty}a_{n}b_{n}x^n$ satisfies $R\ge R_{1}R_{2}$

But for my problem,I can't work this.Thank you

Answer 1

For every $m\geqslant1$ and $n\geqslant1$, one has $1\leqslant\left(1+\frac1{n^{m}}\right)^n\leqslant\left(1+\frac1{n}\right)^n\leqslant3$. Hence, if every $a_k$ is nonnegative, $\alpha\leqslant b_m\leqslant3\alpha$ for every $m\geqslant1$, where $\alpha=\sum\limits_{k=1}^\infty a_k$.

Thus, as soon as every $a_k$ is nonnegative, not every $a_k$ is zero, and the series $\sum\limits_ka_k$ converges, the radius of convergence of the series $\sum\limits_{n=1}^\infty b_mx^m$ is $1$.