1. ## Series, again.

Hi again. I really need full solutions of the following, because it seems that my ways of solving are wrong or just too short.

1) Find the sum of the series. Posted it before, but I just didn't understand how to continue mr fantastic's solution and also I don't know anything about gamma-function, which Krizalid used in his solution. Anwser is 1-ln2, by the way.
$\displaystyle \sum_{n=1}^{\infty}\frac{6^n}{n(n+1)x^n}$ x=12

2) Check if series is convergent. I know that it does not converge but can't prove it.
$\displaystyle \sum_{n=3}^{\infty}\frac{n+1}{\sqrt[3]{n^4}\ln^4(n+1)}$

3) Find all the values of x for which the series converges. By my way of solving itm the answer is $\displaystyle x\in[-3;3)$
$\displaystyle \sum_{n=2}^{\infty}n^4\bigg(\frac{x^3n+4}{27n+\sin (2n)}\bigg)^n$

2. Originally Posted by Rist
Hi again. I really need full solutions of the following, because it seems that my ways of solving are wrong or just too short.

1) Find the sum of the series. Posted it before, but I just didn't understand how to continue mr fantastic's solution and also I don't know anything about gamma-function, which Krizalid used in his solution. Anwser is 1-ln2, by the way.
$\displaystyle \sum_{n=1}^{\infty}\frac{6^n}{n(n+1)x^n}$

2) Check if series is convergent. I know that it does not converge but can't prove it.
$\displaystyle \sum_{n=3}^{\infty}\frac{n+1}{\sqrt[3]{n^4}\ln^4(n+1)}$

3) Find all the values of x for which the series converges. By my way of solving itm the answer is $\displaystyle x\in[-3;3)$
$\displaystyle \sum_{n=2}^{\infty}n^4\bigg(\frac{x^3n+4}{27n+\sin (2n)}\bigg)^n$
I will help with the second and third, but first tell me what you hae done?

The hint for the second is the limit comparison test

and the third's hint is Root test all the way.

3. Hello

Originally Posted by Rist
Hi again. I really need full solutions of the following, because it seems that my ways of solving are wrong or just too short.

1) Find the sum of the series. Posted it before, but I just didn't understand how to continue mr fantastic's solution and also I don't know anything about gamma-function, which Krizalid used in his solution. Anwser is 1-ln2, by the way.
$\displaystyle \sum_{n=1}^{\infty}\frac{6^n}{n(n+1)x^n}$
$\displaystyle \frac{6^n}{n(n+1)x^n}=\left(\frac 6x\right)^n \frac{1}{n(n+1)}=\left(\frac 6x\right)^n \cdot \left(\frac 1n-\frac{1}{n+1}\right)$

The sum is now :

$\displaystyle S=\sum_{n=1}^{\infty} \left(\frac 6x\right)^n \cdot \frac 1n-\sum_{n=1}^{\infty} \left(\frac 6x\right)^n \cdot \frac{1}{n+1}$

--------------------------
Changing the indice of the second one, we get :

$\displaystyle \sum_{n=1}^{\infty} \left(\frac 6x\right)^n \cdot \frac{1}{n+1}=\sum_{n=2}^{\infty} \left(\frac 6x\right)^{n-1} \cdot \frac 1n$.

For the first one, we're going to rewrite in order to start from the same n.

$\displaystyle \sum_{\color{red}n=1}^{\infty} \left(\frac 6x\right)^n \cdot \frac 1n=\frac 6x \cdot \frac 11 +\sum_{\color{red}n=2}^{\infty} \left(\frac 6x\right)^n \cdot \frac 1n$$\displaystyle =\frac 6x+\sum_{n=2}^{\infty} \left(\frac 6x\right)^n \cdot \frac 1n$
--------------------------

$\displaystyle S=\frac 6x+\sum_{n=2}^{\infty} \left(\frac 6x\right)^n \cdot \frac 1n-\sum_{n=2}^{\infty} \left(\frac 6x\right)^{n-1} \cdot \frac 1n$

$\displaystyle S=\frac 6x+\sum_{n=2}^{\infty} \frac 1n \left(\left(\frac 6x\right)^n-\left(\frac 6x\right)^{n-1}\right)$

$\displaystyle S=\frac 6x+\sum_{n=2}^{\infty} \frac 1n \cdot \left(\frac 6x\right)^n \cdot \left(1-\frac x6\right)$

$\displaystyle S=\frac 6x+\left(1-\frac x6\right) \cdot \sum_{n=2}^{\infty} \frac 1n \cdot \left(\frac 6x\right)^n$

Then, recognize the power series :

$\displaystyle -\ln(1-t)=\sum_{n=1}^{\infty} \frac{t^n}{n}=t+\sum_{\color{red}n=2}^{\infty} \frac{t^n}{n}$

$\displaystyle \implies \sum_{\color{red}n=2}^{\infty} \frac{t^n}{n}=-\ln(1-t)-t$

Here, $\displaystyle t=\frac 6x$

So :

$\displaystyle S=\frac 6x+\left(1-\frac x6\right) \cdot \left(-\ln \left(1-\frac 6x\right)-\frac 6x \right)$

$\displaystyle S=\frac 6x-\ln \left(1-\frac 6x\right)-\frac 6x+\frac x6 \cdot \ln \left(1-\frac 6x\right)+1$

$\displaystyle S=-\ln \left(1-\frac 6x\right)+\frac x6 \cdot \ln \left(1-\frac 6x\right)+1$

$\displaystyle S=\ln \left(1-\frac 6x\right) \cdot \left(\frac x6-1\right)+1$

Check it, because I'm quite tired and not really sure

Edit : I just saw that you wanted x=12...

$\displaystyle S(12)=\ln \left(1-\frac 12\right) \cdot (2-1)+1=\ln \left(\frac 12\right)+1=1-\ln 2 \quad \blacksquare$

!!!!!

4. 2nd:

$\displaystyle \sum_{n=3}^{\infty}\frac{n+1}{\sqrt[3]{n^4}\ln^4(n+1)}\sim\sum_{n=3}^{\infty}\frac{n}{n^ \frac{4}{3}\ln^4n}=\sum_{n=3}^{\infty}\frac{1}{n^\ frac{1}{3}\ln^4n}$

I don't know whether it more or less than $\displaystyle \frac{1}{n}$

3rd:

$\displaystyle \sum_{n=2}^{\infty}n^4\bigg(\frac{x^3n+4}{27n+\sin (2n)}\bigg)^n$

For $\displaystyle x\neq0$:

$\displaystyle \sum_{n=2}^{\infty}n^4\bigg(\frac{x^3n+4}{27n+\sin (2n)}\bigg)^n\sim\sum_{n=2}^{\infty}n^4\bigg(\frac {x^3n}{27n}\bigg)^n$

$\displaystyle \lim_{x\to+\infty}\sqrt[n]{|a_{n}|}=\lim_{x\to+\infty}\bigg|n^\frac{4}{n}*\f rac{x^3n}{27n}\bigg|=\bigg|\frac{x^3}{27}\bigg|$

$\displaystyle -3 < x < 3$

For x=3:
$\displaystyle \sum_{n=2}^{\infty}n^4\bigg(\frac{27n+4}{27n+\sin( 2n)}\bigg)^n\sim\sum_{n=2}^{\infty}n^4$
This one divergent

For x=-3:
$\displaystyle \sum_{n=2}^{\infty}n^4\bigg(\frac{-27n+4}{27n+\sin(2n)}\bigg)^n\sim\sum_{n=2}^{\infty }n^4\bigg(\frac{4}{27n}\bigg)^n$
Convergent.

For x=0:
The same as for x=-3.

Answer is $\displaystyle x\in[-3;3)$

5. Originally Posted by Rist
2nd:

$\displaystyle \sum_{n=3}^{\infty}\frac{n+1}{\sqrt[3]{n^4}\ln^4(n+1)}\sim\sum_{n=3}^{\infty}\frac{n}{n^ \frac{4}{3}\ln^4n}=\sum_{n=3}^{\infty}\frac{1}{n^\ frac{1}{3}\ln^4n}$

I don't know whether it more or less than $\displaystyle \frac{1}{n}$
Bertrand Series

At the bottom of the page, there is the result

6. Hi

Otherwise you can show it :

$\displaystyle \frac{\frac{1}{n^{\frac{1}{3}}\ln^4n}}{\frac{1}{n} }=\frac{n^{\frac{2}{3}}}{\ln^4n}\underset{\infty}{ \to}\infty$

This implies that for $\displaystyle n$ sufficiently large $\displaystyle \frac{\frac{1}{n^{\frac{1}{3}}\ln^4n}}{\frac{1}{n} }>1$ thus $\displaystyle \frac{1}{n^{\frac{1}{3}}\ln^4n}>\frac{1}{n}$.

7. Originally Posted by Rist

(...) and also I don't know anything about gamma-function, which Krizalid used in his solution. Anwser is 1-ln2, by the way.
$\displaystyle \sum_{n=1}^{\infty}\frac{6^n}{n(n+1)x^n}$
Let's see another method:

We'll use the Maclaurin expansion for $\displaystyle \ln(1-x)$ and the integral parameter $\displaystyle \frac1{n+1}=\int_0^1x^n\,dx.$ Put these together:

\displaystyle \begin{aligned} \sum\limits_{n\,=\,1}^{\infty }{\frac{1}{n(n+1)2^{n}}}&=\sum\limits_{n\,=\,1}^{\ infty }{\Bigg\{ \frac{1}{n2^{n}}\int_{0}^{1}{x^{n}\,dx} \Bigg\}} \\ & =\int_{0}^{1}{\left\{ \sum\limits_{n\,=\,1}^{\infty }{\frac{1}{n}\bigg( \frac{x}{2} \bigg)^{n}} \right\}\,dx} \\ & =-\int_{0}^{1}{\ln \bigg( 1-\frac{x}{2} \bigg)\,dx}, \end{aligned}

where the last integral can be calculated by standard techniques.

8. What about the 3rd one?

9. Originally Posted by Rist
For the last one we see that the root test gives

$\displaystyle |x|<3$

For $\displaystyle x=3$ try the n-th term test, actually try it for both.

10. $\displaystyle \sum_{n=2}^{\infty}n^4\bigg(\frac{x^3n+4}{27n+\sin (2n)}\bigg)^n\sim\sum_{n=2}^{\infty}n^4\bigg(\frac {4}{27n}\bigg)^n$
It's for all negative x. Is it correct? Because if it is, the answer changes to $\displaystyle x\in(-\infty;3)$

11. Originally Posted by Rist
$\displaystyle \sum_{n=2}^{\infty}n^4\bigg(\frac{x^3n+4}{27n+\sin (2n)}\bigg)^n\sim\sum_{n=2}^{\infty}n^4\bigg(\frac {4}{27n}\bigg)^n$
It's for all negative x. Is it correct? Because if it is, the answer changes to $\displaystyle x\in(-\infty;3)$
You messed up on the asymptotic equivalence, you inexplicably dropped the n in the numerator in the quantity raised to the nth power.