How do you use the direct Comparison test on the infinite series #sum_(n=1)^oo(1+sin(n))/(5^n)# ?

Question

How do you use the direct Comparison test on the infinite series  #sum_(n=1)^oo(1+sin(n))/(5^n)# ?

Samantha Adkison · Accepted Answer

"Using a test" means deciding if the series converges or diverges. If you can compare this series to a larger one that converges, that means this series converges too.

The value of sin(x) is always between -1 and 1; #sin(n) >= -1# means our series is all non-negative terms. And since we have #sin(n) <= 1# for all n, comparing the nth term we get:

#(1+sin(n))/(5^n) < 2/(5^n)#

This means our sum is less than #sum_(n=1)^(oo)2/(5^n)#, which converges as a geometric series with r = 1/5 < 1.

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Isaiah Allen · Answer

undefined To use the Direct Comparison Test on the series $\sum_{n=1}^{\infty} \frac{1+\sin(n)}{5^n}$, we need to find a series whose terms are greater than or equal to the terms of the given series, term by term, and which converges or diverges.

Firstly, note that $|\sin(n)| \leq 1$ for all $n \in \mathbb{N}$. Therefore, $1 + \sin(n) \leq 2$ for all $n$.

Thus, $\frac{1+\sin(n)}{5^n} \leq \frac{2}{5^n}$ for all $n$.

The series $\sum_{n=1}^{\infty} \frac{2}{5^n}$ is a geometric series with $a = 2$ and $r = \frac{1}{5}$. It converges since $|r| < 1$.

So, by the Direct Comparison Test, since $\frac{1+\sin(n)}{5^n} \leq \frac{2}{5^n}$ for all $n$ and $\sum_{n=1}^{\infty} \frac{2}{5^n}$ converges, then $\sum_{n=1}^{\infty} \frac{1+\sin(n)}{5^n}$ converges.