Use the Direct Comparison Test (if possible) to determine whether the series $\sum _ { n = 1 } ^ { \infty } \frac { 2 ^ { n } } { 8 ^ { n } + 1 }$ converges or diverges.
A) diverges
B) converges

Question

Quizplus · Accepted Answer

Let us take $a_n=\frac{2^n}{8^n+1}$. Also, for all $n\geq1$ we have $a_n>0$. Then we take $b_n=\frac{1}{8^n}$. We have
$$
\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{2^n}{8^n+1}\cdot8^n=\lim_{n\to\infty}\frac{2^n}{1+8^{-n}}=+\infty,
$$
where we used that $\lim_{n\to\infty} 8^{-n}=0$. Hence we conclude that $\sum b_n=\sum\frac{1}{8^n}=1/7<+\infty$ and using the Direct Comparison Test, we get that $\sum a_n$ also converges.

Use the Direct Comparison Test (If Possible) to Determine Whether ∑n=1∞2n8n+1\sum _ { n = 1 } ^ { \infty } \frac { 2 ^ { n } } { 8 ^ { n } + 1 }∑n=1∞​8n+12n​

Use the Direct Comparison Test (If Possible) to Determine Whether $\sum _ { n = 1 } ^ { \infty } \frac { 2 ^ { n } } { 8 ^ { n } + 1 }$