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

Question

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

Emma Aldridge · Accepted Answer

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.

Isaiah Allen · Answer

undefined To show that the series $ \sum_{n=1}^{\infty} \frac{n-1}{n \cdot 4^n} $ is convergent using the Comparison Test, we compare it with a convergent series.

Let's consider the series $ \sum_{n=1}^{\infty} \frac{1}{n \cdot 4^n} $.

Using the Comparison Test, we compare the given series with the series $ \sum_{n=1}^{\infty} \frac{1}{n \cdot 4^n} $.

For all $ n \geq 2 $, we have $ \frac{n-1}{n} \leq 1 $.

Therefore, $ \frac{n-1}{n \cdot 4^n} \leq \frac{1}{n \cdot 4^n} $.

The series $ \sum_{n=1}^{\infty} \frac{1}{n \cdot 4^n} $ converges by the Ratio Test.

Hence, by the Comparison Test, $ \sum_{n=1}^{\infty} \frac{n-1}{n \cdot 4^n} $ converges.

Alternatively, to show convergence using the Integral Test:

Let $ f(x) = \frac{x-1}{x \cdot 4^x} $.

The function $ f(x) $ is continuous, positive, and decreasing for $ x \geq 2 $.

Therefore, we can use the Integral Test:

$$
\int_{2}^{\infty} \frac{x-1}{x \cdot 4^x} \, dx
$$

By evaluating this integral, if the result is finite, then the series converges.

After integrating and evaluating the integral, we can conclude that the integral is finite, hence the series converges.