0
$\begingroup$

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.

$\endgroup$
  • 3
    $\begingroup$ For $k\geq 4$, $\frac{1}{2}+\frac{1}{k}\leq\frac{3}{4}$. Now, compare to a geometric series. $\endgroup$ – Michael Burr Feb 6 '18 at 12:36
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Why not use $k = 3$? $\endgroup$ – Mr. Chip Feb 6 '18 at 12:39
  • $\begingroup$ @Mr.Chip No reason, I just picked a value. $\endgroup$ – Michael Burr Feb 6 '18 at 12:41
  • $\begingroup$ @DanAntonio's answer is simpler than mine, please take a look. $\endgroup$ – Michael Burr Feb 6 '18 at 12:43
7
$\begingroup$

Directly with the $\;n\,-$ th root test:

$$\sqrt[n]{\left(\frac12+\frac1n\right)^n}=\frac12+\frac1n\xrightarrow[n\to\infty]{}\frac12<1$$

and thus the series converges.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.