1. ## test for convergence

test for convergence: sigma from n=1 to infinitie of [1-(1/squarerootn)]^n

2. screams comparison test to me. think famous limits. however im pretty rubbish at analysis.

3. Originally Posted by twilightstr

test for convergence: $\sum_{n=1}^{\infty} \left(1- \frac{1}{\sqrt{n}} \right)^n$
applying the mean value theorem to the function $f(x)=\ln x, \ 0 < a < b,$ we'll get: $\ln b - \ln a > \frac{b-a}{b}. \ \ \ \ \ (1)$

now suppose $n \geq 2$ and in $(1)$ put $a=\sqrt{n} - 1$ and $b=\sqrt{n}$ to get $\ln \sqrt{n} - \ln(\sqrt{n} - 1) > \frac{1}{\sqrt{n}}.$ therefore:

$\left(1 - \frac{1}{\sqrt{n}} \right)^n < e^{-\sqrt{n}}$ and hence your series is convergent by the comparison test.

(note that $\lim_{n\to\infty} \frac{e^{-\sqrt{n}}}{\frac{1}{n^2}}=\lim_{x\to\infty} \frac{x^4}{e^x}=0$ and thus the series $\sum e^{-\sqrt{n}}$ is convergent by the limit comparison test.)

4. Originally Posted by NonCommAlg
applying the mean value theorem to the function $f(x)=\ln x, \ 0 < a < b,$ we'll get: $\ln b - \ln a > \frac{b-a}{b}. \ \ \ \ \ (1)$

now suppose $n \geq 2$ and in $(1)$ put $a=\sqrt{n} - 1$ and $b=\sqrt{n}$ to get $\ln \sqrt{n} - \ln(\sqrt{n} - 1) > \frac{1}{\sqrt{n}}.$ therefore:

$\left(1 - \frac{1}{\sqrt{n}} \right)^n < e^{-\sqrt{n}}$ and hence your series is convergent by the comparison test.

(note that $\lim_{n\to\infty} \frac{e^{-\sqrt{n}}}{\frac{1}{n^2}}=\lim_{x\to\infty} \frac{x^4}{e^x}=0$ and thus the series $\sum e^{-\sqrt{n}}$ is convergent by the limit comparison test.)

5. Originally Posted by NonCommAlg
applying the mean value theorem to the function $f(x)=\ln x, \ 0 < a < b,$ we'll get: $\ln b - \ln a > \frac{b-a}{b}. \ \ \ \ \ (1)$

now suppose $n \geq 2$ and in $(1)$ put $a=\sqrt{n} - 1$ and $b=\sqrt{n}$ to get $\ln \sqrt{n} - \ln(\sqrt{n} - 1) > \frac{1}{\sqrt{n}}.$ therefore:

$\left(1 - \frac{1}{\sqrt{n}} \right)^n < e^{-\sqrt{n}}$ and hence your series is convergent by the comparison test.
Just a quick remark: we can also see this as the usual convexity inequality $1-x\leq e^{-x}$ for $x=\frac{1}{\sqrt{n}}$, then raised to the power $n$.