How do you use the comparison test (or the limit comparison test) for $\frac{1 + sin (x)}{10^{x}}$ $(1+sin(x))/10^x$ ?

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Answer 1 · 2015-10-15T13:56:00

For the series $\infty \sum i = 1 \frac{1 + sin (n)}{10^{n}}$ $sum_(i=1)^oo (1+sin(n))/10^n$ use $1 + sin n \leq 2$ $1+sinn <= 2$ .

So $\frac{1 + sin (n)}{10^{n}} \leq \frac{2}{10^{n}}$ $(1+sin(n))/10^n <= 2/10^n$

$\infty \sum i = 1 \frac{2}{10^{n}}$ $sum_(i=1)^oo 2/10^n$ is a geometric series with $r = \frac{1}{10}$ $r = 1/10$ so it converges.

How do you use the comparison test (or the limit comparison test) for (1+sin(x))/10^x1+sin(x)10x(1+sin(x))/10^x?