# show that $\sum_{k=1}^{\infty}{\left ( \frac{1}{2} + \frac{1}{k}\right )^k}$ converges

How can I show that $\sum_{k=1}^{\infty}{\left ( \frac{1}{2} + \frac{1}{k}\right )^k}$ converges ? I really have no idea how to start.

• For $k\geq 4$, $\frac{1}{2}+\frac{1}{k}\leq\frac{3}{4}$. Now, compare to a geometric series. – Michael Burr Feb 6 '18 at 12:36

Observe that when $k\geq 4$, we know that $\frac{1}{k}\leq\frac{1}{4}$. Therefore, $$\left(\frac{1}{2}+\frac{1}{k}\right)\leq\frac{3}{4}.$$ Therefore, $$\sum_{k=4}^\infty \left(\frac{1}{2}+\frac{1}{k}\right)^k\leq \sum_{k=4}^\infty\left(\frac{3}{4}\right)^k$$ The RHS is a geometric series which converges since $\frac{3}{4}<1$. Therefore, by the comparison test, the LHS converges. Adding on the first few terms doesn't change convergence.
• Why not use $k = 3$? – Mr. Chip Feb 6 '18 at 12:39
Directly with the $\;n\,-$ th root test:
$$\sqrt[n]{\left(\frac12+\frac1n\right)^n}=\frac12+\frac1n\xrightarrow[n\to\infty]{}\frac12<1$$