Show that the series $\sum_{k=1}^{\infty} k^na_k$ converges $\forall n \in \mathbb{N}$

Question

Problem : Suppose that the series $\sum_{}^{} a_k$ of positive real numbers converges by virtue of the root test or ratio test. Show that the series $\sum_{k=1}^{\infty} k^na_k$ converges $\forall n \in \mathbb{N}$

My Proof :

Let $n \in \mathbb{N}$ be given.

Suppose $\sum_{}^{} a_k$ converges by ratio test, then there exists $\limsup_{k\to\infty}\ \frac{a_{k+1}}{a_k}$

Let $R=\limsup_{k\to\infty}\ \frac{a_{k+1}}{a_k}$ then $R<1$

Choose $c$ s.t. $R<c<1$ then, there exists $n_0 \in \mathbb{N}$ s.t. $ \frac{a_{k+1}}{a_k}<c\ \ \ \forall k >n_0$

Let $M=\frac{a_{n_0}}{c^{n_0}}$ then $a_k \le Mc^k\ \ \ \forall k >n_0$

which implies $k^na_k \le \frac{Mk^n}{c^{-k}}\ \ \ \forall k >n_0$

Since $\frac{1}{c}>1$, $\sum_{}^{} \frac{Mk^n}{c^{-k}}$ converges.

Since sequence $k^na_k$ is positive, $\sum_{k=1}^{\infty} k^na_k$ converges by comparison test.

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But I realized later convergence of seires by ratio test doesn't imply $R<1$ (because there exists convergent series if R=1)

So, I think my proof is valid only R<1 (not R=1)

So.. now, what should I do for verifying when R=1??

Answer 1

While some series may converge even if $R=1$, in that case they do not converge "by virtue of the ratio test". In that case the ratio test is inconclusive and some other method must be used to determine convergence/divergence. Since you are assuming the ratio test here IS conclusive, that means $R<1$, so your proof is correct.