Question

How do you show that `sum(n-1)/(n*4^n)` is convergent using the Comparison Test or Integral Test?

Answer 1

It's simplest to use the comparison test. Let `a_{n}=\frac{n-1}{n\cdot 4^{n}}` and let `b_{n}=\frac{1}{4^{n}}`. Note that `0\leq a_{n}\leq b_{n}` for all integers `n\geq 1` since `\frac{n-1}{n}\leq 1` for all integers `n\geq 1`.

Also note that `\sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}(\frac{1}{4})^{n}` converges since it is geometric with common ratio `r=1/4`, which satisfies `|r|<1` (in fact, it converges to `\frac{1/4}{1-1/4}=\frac{1}{3}`).

The Comparison Test can now be used to say that `\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}\frac{n-1}{n\cdot 4^{n}}` converges.