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

1 Answer
Bill K.
May 29, 2015

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.

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