test the series for convergence or divergence. ∑(n = 1 to ∞) (1 / n^7 + 1 / 7^n) convergent divergent need…

Explanation:

Step1: Recall series - convergence rules

A series $\sum_{n = 1}^{\infty}(a_n + b_n)=\sum_{n = 1}^{\infty}a_n+\sum_{n = 1}^{\infty}b_n$ if both $\sum_{n = 1}^{\infty}a_n$ and $\sum_{n = 1}^{\infty}b_n$ converge.

Step2: Test $\sum_{n = 1}^{\infty}\frac{1}{n^7}$

By the p - series test, for a series $\sum_{n = 1}^{\infty}\frac{1}{n^p}$, if $p>1$, the series converges. Here $p = 7>1$, so $\sum_{n = 1}^{\infty}\frac{1}{n^7}$ converges.

Step3: Test $\sum_{n = 1}^{\infty}\frac{1}{7^n}$

This is a geometric series of the form $\sum_{n = 1}^{\infty}ar^{n - 1}$ (rewrite $\sum_{n = 1}^{\infty}\frac{1}{7^n}=\sum_{n = 1}^{\infty}\frac{1}{7}\times(\frac{1}{7})^{n - 1}$), where $a=\frac{1}{7}$ and $r=\frac{1}{7}$. For a geometric series $\sum_{n = 1}^{\infty}ar^{n - 1}$, if $|r|<1$, the series converges. Since $|r|=\frac{1}{7}<1$, $\sum_{n = 1}^{\infty}\frac{1}{7^n}$ converges.

Step4: Determine convergence of the sum series

Since $\sum_{n = 1}^{\infty}\frac{1}{n^7}$ and $\sum_{n = 1}^{\infty}\frac{1}{7^n}$ both converge, then $\sum_{n = 1}^{\infty}(\frac{1}{n^7}+\frac{1}{7^n})=\sum_{n = 1}^{\infty}\frac{1}{n^7}+\sum_{n = 1}^{\infty}\frac{1}{7^n}$ converges.

Answer:

convergent