# True or False? If a series $\sum_{n=1}^{\infty} a_k$ converges, then $\sum_{n=1}^{\infty} k^pa_k$ converges.

True or False? If a series $\sum_{n=1}^{\infty} a_k$ converges by either the ratio or root test, and if p > 0 is any constant, then $$\sum_{n=1}^{\infty} k^pa_k$$ converges by the same test.

My instinct is that this is true, but I am unsure as to how to go about proving it.

• What if $a_k=\frac1{2^k}$ and $k=2$? – Gregory Grant Feb 16 '16 at 2:20
• What is $k$? What is $n$? Shouldn't $k$ be $n$? – Clement C. Feb 16 '16 at 2:21
• @ClementC. obviously a typo – Gregory Grant Feb 16 '16 at 2:21
• @GregoryGrant Yes, but I also cannot (honestly) parse the part of the question saying " then k=1 ak." (and your comment -- $p=2$, right? :) – Clement C. Feb 16 '16 at 2:22
• @GregoryGrant Please fix your "obvious" typo - it's not obvious what you meant to say. – Zubin Mukerjee Feb 16 '16 at 2:54

Hints: (1) What is $\lim_{k\to\infty}\frac{(k+1)^p}{k^p}$? (2) What is $\lim_{k\to\infty} (k^p)^{1/k}$?
• I am assuming we are using the simple version of the Ratio Test.If you are using the $\limsup$ version, some small changes will have to be made. We are told that $\lim\left|\frac{a_{k+1}}{a_k}\right|\lt1$. We want to show that $\lim\left|\frac{(k+1)^pa_{k+1}}{k^pa_k}\right|\lt 1$. Since $\lim \frac{(k+1)^p}{k^p}=1$, what we want to show follows. Root Test is similar. – André Nicolas Feb 16 '16 at 2:54