How can I prove that $\sum ( 1 - \frac{2}{\sqrt{n}} )^n$ converges?
I've tried root test, and ratio test, but it was inconclusive.
Wolfram Alpha also says that this sum is convergent by comparison test.
What series can I use for comparison test?
Asked
Active
Viewed 132 times
2
Integrand
- 8,457
evelynn0602
- 21
-
Hint: For $n\geq 5$, $n\log \left( {1 - \frac{2}{{\sqrt n }}} \right) < n\left( { - \frac{2}{{\sqrt n }}} \right) = - 2\sqrt n$, because $\log(1+x)<x$ for all $x>-1$. – Gary Mar 26 '20 at 11:47
2 Answers
2
$$\left(1-\frac{2}{\sqrt{n}}\right)^n=\left(\left(1-\frac{2}{\sqrt{n}}\right)^{-\frac{\sqrt{n}}{2}}\right)^{-2\sqrt{n}}\sim\frac{1}{e^{2\sqrt{n}}}<\frac{1}{\frac{\left(2\sqrt{n}\right)^4}{4!}}$$
Michael Rozenberg
- 194,933
0
Another way to see this series converges is by Schlömilch's Test, a generalization of the Cauchy Condensation Test. Basically, if $u(n)$ is a positive increasing sequence with a uniformly bounded ratio of successive differences, then $\sum_{n} f(n)$ converges iff $\sum_{n} (u(n+1)-u(n)) f(u(n))$ converges. In our case, put $u(n)=n^2$: we have $u(n+1)/u(n)<5$ and the new series is $$ \sum (2n+1)\left(1-\frac{2}{n}\right)^{n^2} $$Now we can use the Root Test: $$ \lim_{n\to\infty}\left|\sqrt[n]{(2n+1)\left(1-\frac{2}{n}\right)^{n^2}}\right| $$We will split this into two limits. $$ =\lim_{n\to\infty}\left|\sqrt[n]{2n+1}\right|\cdot \lim_{n\to\infty}\left|\left(1-\frac{2}{n}\right)^{n}\right| $$These limits are both well-known: the first is $1$ and the second is $e^{-2}$. Their product is less than $1$, so we have convergence.
Integrand
- 8,457