1. Power series and convergence

Could somebody explain me how to find the convergence and the radius of the convergence for the following power series?

2. Did you try the ratio test. When you do and take the limit,

$\frac{a_{n+1}}{a_{n}}=\frac{ln(k+2)(x+1)^{k+1}}{k+ 2}\cdot\frac{k+1}{ln(k+1)(x+1)^{k}}$

$\lim_{k\rightarrow{\infty}}\frac{a_{n+1}}{a_{n}}=| x+1|$

I may as well finish.

$-1

$-2

Diverges if x<-2 or of x>0

Check the endpoints:

If we enter in x=-2 into our series we get $\sum_{k=1}^{\infty}\frac{(-1)^{k}ln(k+1)}{k+1}$, converges. (converges to $\approx{-.15986...}$)

If we enter in x=0, we get $\sum_{k=1}^{\infty}\frac{ln(k+1)}{k+1}$, diverges.

So, the interval of convergence is [-2,0) with radius of 1

My assessment agrees with MS except are you sure when x=0 it converges?.

3. Originally Posted by totalnewbie
Could somebody explain me how to find the convergence and the radius of the convergence for the following power series?
Galactus's method is totally right, but just so you get more than one method

Root test

$\sum_{n=1}^{\infty}\frac{\ln(n+1)(x+1)^n}{n+1}$

So we need to find all values of x such that

$\lim_{n\to\infty}\bigg|\frac{\ln(n+1)(x+1)^n}{n+1} \bigg|^{\frac{1}{n}}<1$

So we see that

$\lim_{n\to\infty}\bigg|\frac{\ln^{\frac{1}{n}}(n+1 )(x+1)}{(n+1)^{\frac{1}{n}}}\bigg|<1\Rightarrow|x+ 1|<1$

Or in other words

$x+1<1\Rightarrow{x<0}$

or

$x+1>-1\Rightarrow{x>-2}$

Now we need to check endpoint convergence

at $x=0$ we have

divergence by integral test

at $x=-2$ we have

$\sum_{n=0}^{\infty}\frac{ln(n+1)(-1)^n}{n+1}$

and since $\exists{N}\backepsilon\forall{n>N},a_{n+1}

and also since $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{\ln(n+ 1)}{n+1}=0$

this series converges by AST

Therefoe IOC is $[-2,0)$

EDIT: For purely supplemetary purposes

Let $L=\lim_{n\to\infty}\ln^{\frac{1}{n}}(n+1)\Rightarr ow{\ln(L)=\lim_{n\to\infty}\frac{\ln(\ln(n+1))}{n} }$

For the right hand limit you can use many methods, L'hopital's is most preferable here

$\lim_{n\to\infty}\frac{\ln(\ln(n+1))}{n}=\lim_{n\t o\infty}\frac{\frac{1}{(n+1)\ln(n+1)}}{1}=0$

So we have that

$\ln(L)=\lim_{n\to\infty}\frac{\ln(\ln(n+1))}{n}=0\ Rightarrow{L=e^0=1}$

Use the same method to compute $L=\lim_{n\to\infty}(n+1)^{\frac{1}{n}}$

4. Originally Posted by galactus
Did you try the ratio test. When you do and take the limit,

$\frac{a_{n+1}}{a_{n}}=\frac{ln(k+2)(x+1)^{k+1}}{k+ 2}\cdot\frac{k+1}{ln(k+1)(x+1)^{k}}$

$\lim_{k\rightarrow{\infty}}\frac{a_{n+1}}{a_{n}}=| x+1|$

I may as well finish.

$-1

$-2

Diverges if x<-2 or of x>0

Check the endpoints:

If we enter in x=-2 into our series we get $\sum_{k=1}^{\infty}\frac{(-1)^{k}ln(k+1)}{k+1}$, converges. (converges to $\approx{-.15986...}$)

If we enter in x=0, we get $\frac{ln(k+1)}{k+1}$, diverges.

So, the interval of convergence is [-2,0) with radius of 1

My assessment agrees with MS except are you sure when x=0 it converges?.
Thanks so much! I fixed it before you posted that , the entire time I kept typing x instead of x+1, therefore when I saw x=0 I said a sequence of all zero terms, I am sorry for my stupid error , I knew it would bite me sooner or later

Just to show once again, now that I see it is $(x+1)^n$ and not $x^n$

x=0 gives

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

and as we stated earlier the existence of eventual monotonic decrease and all the terms are positve we can apply the integral test

$\int_1^{\infty}\frac{ln(n+1)}{n+1}=\frac{1}{2}\ln^ 2(n+1)\bigg|_1^{\infty}=\infty$

therefore divergent