# Math Help - Test for convergence

1. ## Test for convergence

$\displaystyle \sum_{n = 1}^{\infty} \displaystyle \frac{(-1)^n}{ln(1+e^n)}$

$\displaystyle \sum_{n = 1}^{\infty} (-1)^n (1-\displaystyle \frac{1}{n})^{n^2}$

known that the first one is conditionally convergent and the second one is absolutely convergent, which test shud i use to prove it

2. Originally Posted by wopashui
$\displaystyle \sum_{n = 1}^{\infty} \displaystyle \frac{(-1)^n}{ln(1+e^n)}$

$\displaystyle \sum_{n = 1}^{\infty} (-1)^n (1-\displaystyle \frac{1}{n})^{n^2}$

known that the first one is conditionally convergent and the second one is absolutely convergent, which test shud i use to prove it

$\displaystyle{\frac{1}{\ln(1+e^n)}$ is monotonically descending convergent to zero so $\displaystyle \sum_{n = 1}^{\infty} \displaystyle \frac{(-1)^n}{\ln(1+e^n)}}$ is cond. convergent.

It isn't abs. convergent since $\displaystyle{\frac{1}{\ln(1+e^n)}\geq\frac{1}{\ln (2e^n)}=\frac{1}{\ln 2+n}$ , which is divergent.

For the second one the root test gives $\displaystyle{\left[\left(1-\frac{1}{n}\right)^{n^2}\right]^{1/n}=\left(1-\frac{1}{n}\right)^n\xrightarrow [n\to\infty]{}e^{-1}<1$ ...

Tonio

3. Originally Posted by tonio
$\displaystyle{\frac{1}{\ln(1+e^n)}$ is monotonically descending convergent to zero so $\displaystyle \sum_{n = 1}^{\infty} \displaystyle \frac{(-1)^n}{\ln(1+e^n)}}$ is cond. convergent.

It isn't abs. convergent since $\displaystyle{\frac{1}{\ln(1+e^n)}\geq\frac{1}{\ln (2e^n)}=\frac{1}{\ln 2+n}$ , which is divergent.

For the second one the root test gives $\displaystyle{\left[\left(1-\frac{1}{n}\right)^{n^2}\right]^{1/n}=\left(1-\frac{1}{n}\right)^n\xrightarrow [n\to\infty]{}e^{-1}<1$ ...

Tonio
do you use the alternating series test for the first one, I dun quite understand how it works, I understanf why is not absolutely convergent, but why is convergent without the absolute value sign?

and can you explain why $\frac{1}{\ln 2+n}$ is diverges, it seems going to 0 as n goes to infinty

4. Originally Posted by wopashui
do you use the alternating series test for the first one, I dun quite understand how it works, I understanf why is not absolutely convergent, but why is convergent without the absolute value sign?

and can you explain why $\frac{1}{\ln 2+n}$ is diverges, it seems going to 0 as n goes to infinty

The sequence does, but not the series formed with it, which is what we were talking about.

Tonio

5. Originally Posted by tonio
$\displaystyle{\frac{1}{\ln(1+e^n)}$ is monotonically descending convergent to zero so $\displaystyle \sum_{n = 1}^{\infty} \displaystyle \frac{(-1)^n}{\ln(1+e^n)}}$ is cond. convergent.

It isn't abs. convergent since $\displaystyle{\frac{1}{\ln(1+e^n)}\geq\frac{1}{\ln (2e^n)}=\frac{1}{\ln 2+n}$ , which is divergent.

For the second one the root test gives $\displaystyle{\left[\left(1-\frac{1}{n}\right)^{n^2}\right]^{1/n}=\left(1-\frac{1}{n}\right)^n\xrightarrow [n\to\infty]{}e^{-1}<1$ ...

Tonio
so you said ] $\displaystyle{\frac{1}{\ln(1+e^n)}$ converges to zero then you said it is diverges, so is it diverges or converges? and why is $\frac{1}{\ln 2+n}$ divers, it seems approaching to 0 as n goes to infinty

6. Originally Posted by wopashui
so you said ] $\displaystyle{\frac{1}{\ln(1+e^n)}$ converges to zero then you said it is diverges, so is it diverges or converges? and why is $\frac{1}{\ln 2+n}$ divers, it seems approaching to 0 as n goes to infinty

You continue to confuse infinite series with the infinite sequence it is formed with!

One thing is a series $\sum\limits^\infty_{n=1}a_n$ , and another one is its sequence $\left\{a_n\right\}^\infty_{n=1}$ ...

Tonio

7. Originally Posted by tonio
You continue to confuse infinite series with the infinite sequence it is formed with!

One thing is a series $\sum\limits^\infty_{n=1}a_n$ , and another one is its sequence $\left\{a_n\right\}^\infty_{n=1}$ ...

Tonio
so, the series 1/ln2+n >1/2n which is the multiple of the harmonic series, so the series diverges, right?