Please check this convergence test

[ognik]

$ \sum_{n} \ln\left({1+\frac{1}{n}}\right) $

$ \ln\left({1+\frac{1}{n}}\right) = \ln\left({1}\right) + \ln\left({\frac{1}{n}}\right) = 0 +\ln\left({{n}^{-1}}\right) = -\ln\left({n}\right)$

Now $\lim_{{n}\to{\infty}} -\ln\left({n}\right) \ne 0$, therefore the series diverges.

(Also can you suggest an alternate approach?)

 

[GJA]

Hey ognik,

Need to be a bit more careful: $\ln\left(1+\frac{1}{n}\right)\neq \ln(1)+\ln\left(\frac{1}{n}\right)$

 

[ognik]

Oops, blush...

probably because everything else I've tried didn't go anywhere, a hint please? I expect it to diverge because 1/n diverges, so ln 1/n will also diverge, albeit slower ...

 

[Greg]

If you try and calculate the sum, you'll find it's a telescoping series with log(n+1) as the "unresolved" term. The sum is therefore equivalent to log(n + 1) in the limit, so it diverges.

 

[ognik]

Thanks - is there an easy/reliable way to tell if a series is telescoping? (other than writing out the terms and seeing what is left...)

The book reckons $ln (1+\frac{1}{n}) \approx \frac{1}{n}$ - but I don't know why this is, could someone please walk me through it?

 

[MarkFL]

...The book reckons $ln (1+\frac{1}{n}) \approx \frac{1}{n}$ - but I don't know why this is, could someone please walk me through it?

Look at first term of the Maclaurin series:

$$\ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k+1}\frac{x^{k}}{k}$$ where $|x|<1$.

 

[Prove It]

$ \sum_{n} \ln\left({1+\frac{1}{n}}\right) $

$ \ln\left({1+\frac{1}{n}}\right) = \ln\left({1}\right) + \ln\left({\frac{1}{n}}\right) = 0 +\ln\left({{n}^{-1}}\right) = -\ln\left({n}\right)$

Now $\lim_{{n}\to{\infty}} -\ln\left({n}\right) \ne 0$, therefore the series diverges.

(Also can you suggest an alternate approach?)

$\displaystyle \begin{align*} \ln{ \left( 1 + \frac{1}{n} \right) } = \ln{ \left( \frac{n + 1}{n} \right) } = \ln{ \left( n + 1 \right) } - \ln{ \left( n \right) } \end{align*}$

The telescopic nature of this series should now be obvious.

 

[ognik]

All clear thanks.

Could one also argue that ln(n+1) - ln(n) is clearly increasing (because of the shape of ln(n)) and therefore it diverges?

 

[Greg]

Differentiate log(n + 1) - log(n) w.r.t. n. What do you find?

 

[ognik]

Thanks. Do you mean that the slope is > 0, therefore the function diverges?

 

[Greg]

Did you calculate the derivative of log(n + 1) - log(n) w.r.t. n?

 

[ognik]

I got $\frac{-1}{n(n+1)}$ which is always decreasing because n > 0

Oh I see now, typo, > should have been < in my prev. post, apologies.

However, my understanding is that showing a series is decreasing is not enough to decide if it converges to a finite value? (and I already know this series diverges)

 

[Greg]

Right; I believe that's what they call "necessary but not sufficient".