Answer
Converges
Work Step by Step
Th. 9.13 Limit Comparison Test:$\\ $If $a_{n} > 0, b_{n} > 0$, and $\displaystyle \lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=L, \\ $where $L$ is finite and positive, then$\displaystyle \\ \sum_{n=1}^{\infty}a_{n}$ and $\displaystyle \sum_{n=1}^{\infty}b_{n}$ either both converge or both diverge.
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Let $a_{n}=\displaystyle \frac{2^{n}+1}{5^{n}+1}$.
...If the numerator $and$ denominator had the same leading power $(10^{n})$,
we could easily evaluate the limit ...
This is why we observe $b_{n}=\displaystyle \frac{2^{n}}{5^{n}}$
$\displaystyle \sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}(\frac{2}{5})^{n}$
which is a convergent geometric series.
Test the hypothesis of the theorem:
$\displaystyle \lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}$ =$\displaystyle \lim_{n\rightarrow\infty}\frac{\frac{2^{n}+1}{5^{n}+1}}{\frac{2^{n}}{5^{n}}}=\lim_{n\rightarrow\infty}\frac{5^{n}(2^{n}+1)}{2^{n}(5^{n}+1)}$=
$\displaystyle \lim_{n\rightarrow\infty}\frac{10^{n}+5^{n}}{10^{n}+2^{n}}$=
... divide both numerator and denominator with $ 10^{n}$ ...
$=\displaystyle \lim_{n\rightarrow\infty}\frac{1+\frac{1}{2^{n}}}{1+\frac{1}{5^{n}}}=1=L$ (finite and positive)
So,
$\displaystyle \sum_{n=1}^{\infty}a_{n}$ converges as well (since they both converge).
