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• Nov 14th 2008, 07:48 AM
Milus
Can someone help me to sove these.

Find radii of convergence of following series:
1. n(logn)x^n ( n from 1 to infinity)
2.(logn)^(logn )*x^n (n from 2 to infinity)
3. (x^n)/(n^n(1/2)) n from 1 to infinity
4. (x^n)/(n^(1/2))^n n from 1 to infinity
5.((n^n)(x^n))/n! n from 1 to infinity
• Nov 14th 2008, 12:04 PM
Mathstud28
Quote:

Originally Posted by Milus
Can someone help me to sove these.

Find radii of convergence of following series:
1. n(logn)x^n ( n from 1 to infinity)
2.(logn)^(logn )*x^n (n from 2 to infinity)
3. (x^n)/(n^n(1/2)) n from 1 to infinity
4. (x^n)/(n^(1/2))^n n from 1 to infinity
5.((n^n)(x^n))/n! n from 1 to infinity

1. $\displaystyle \sum_{n=1}^{\infty}n\ln(n)x^n$

Note the obvious, the only thing that will dominate this series so that it converges is a geometric term. Therefore $\displaystyle |x|<1$

2.$\displaystyle \sum_{n=2}^{\infty}\ln(n)^{\ln(n)}x^n$

Apply Root test. First noting that $\displaystyle \ln(n)^{\ln(n)}=e^{\ln(n)\ln(\ln(n))}$ and the continuity of the exponential function. The limit tends to x.

3. Note that $\displaystyle \sum_{n=1}^{\infty}\frac{x^n}{n^{\frac{n}{2}}}\leq \sum_{n=1}^{\infty}\frac{x^n}{n!}=e^x-1$

Which converges for all x.

4. I don't see how this is differnt than number three?

5. Using the Root test so we have that

\displaystyle \begin{aligned}\lim_{n\to\infty}\sqrt[n]{\frac{n^nx^n}{n!}}=&\lim_{n\to\infty}\frac{nx}{n! ^{\frac{1}{n}}}\\ &\sim\lim_{n\to\infty}\frac{nx}{(2\pi{n})^{\frac{1 }{2n}}ne^{-1}}\\ &=e|x|<1 \end{aligned}

I leave the endpoints up to you.
• Nov 14th 2008, 02:51 PM
Milus
Thanks.But still I do not understand why do you change logn to lnn.
Also the problem that I did not write correctely is :

(x^n)/(n^(n^(1/2)))
• Nov 14th 2008, 02:56 PM
Mathstud28
Quote:

Originally Posted by Milus
Thanks.But still I do not understand why do you change logn to lnn.
Also the problem that I did not write correctely is :

(x^n)/(n^(n^(1/2)))

Because in upper mathematics $\displaystyle \log(x)=\ln(x)$. It is usually only lower mathematics that I have seen the convention that $\displaystyle \log(x)=\log_{10}(x)$. Regardless, the answer will not change for either.

For $\displaystyle \sum_{n=1}^{\infty}\frac{x^n}{n^{n^{\frac{1}{2}}}}$ use the root test.
• Nov 14th 2008, 08:03 PM
Milus
How to justify step by stem #1,and also that missing #4 that is (x^n)/(n^(n^(1/2))) .
What to use?Help!!!
• Nov 15th 2008, 07:43 AM
Mathstud28
Quote:

Originally Posted by Milus
How to justify step by stem #1,and also that missing #4 that is (x^n)/(n^(n^(1/2))) .
What to use?Help!!!

For number one, you can use the Root or Ratio test to deduce it yourself. Or assume that what I said was correct, and then use the limit comparison test between your series and a geometric series.

As for your second quetsion, I covered that in my last post.
• Nov 16th 2008, 08:30 PM
Milus
(Crying)Hi
I feel really stupid.You practicaly solved that for me and I still do not know how to solve almost any of it.I do not know in what point to use what test and how to find the end points.I am trying over and over and i just cannot do any progress.I have about 10 more similar problems that I have to solve, but I am not even able to solve these here.Please can someone show me step by step solution,so I will be able to understand.
• Nov 17th 2008, 05:55 AM
Milus
I really need help.I am frustrated and I start to give up.
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
• Nov 17th 2008, 12:43 PM
Mathstud28
Quote:

Originally Posted by Mathstud28
1. $\displaystyle \sum_{n=1}^{\infty}n\ln(n)x^n$

Note the obvious, the only thing that will dominate this series so that it converges is a geometric term. Therefore $\displaystyle |x|<1$

The Root Ratio test will give

\displaystyle \color{red}\begin{aligned}\lim_{n\to\infty}\left|\ frac{(n+1)\ln(n+1)x^{n+1}}{n\ln(n)x^n}\right|&=|x| \lim_{n\to\infty}\left|\frac{(n+1)\ln(n+1)}{n\ln(n )}\right|\\ &=|x|\cdot{1}\\ &=|x|<1 \end{aligned}

2.$\displaystyle \sum_{n=2}^{\infty}\ln(n)^{\ln(n)}x^n$

Apply Root test. First noting that $\displaystyle \ln(n)^{\ln(n)}=e^{\ln(n)\ln(\ln(n))}$ and the continuity of the exponential function. The limit tends to x.

After noting what I said above we just have to note that

\displaystyle \color{red}\begin{aligned}\lim_{n\to\infty}e^{\fra c{\ln(n)\ln(\ln(n))}{n}}&=e^{\lim_{n\to\infty}\fra c{\ln(n)\ln(\ln(n))}{n}}\\ &=e^0\\ &=1 \end{aligned}

So \displaystyle \color{red}\begin{aligned}\lim_{n\to\infty}\sqrt[n]{x^ne^{\ln(n)\ln(\ln(n))}}&=|x|\lim_{n\to\infty}e^ {\frac{\ln(n)\ln(\ln(n))}{n}}\\ &=|x|<1 \end{aligned}

3. Note that $\displaystyle \sum_{n=1}^{\infty}\frac{x^n}{n^{\frac{n}{2}}}\leq \sum_{n=1}^{\infty}\frac{x^n}{n!}=e^x-1$

Which converges for all x.

$\displaystyle \color{red}\text{I think this one is obvious, if not say so}$

Or hey just use the Root test if its not obvious

\displaystyle \color{red}\begin{aligned}\lim_{n\to\infty}\sqrt[n]{\frac{x^n}{n^{\frac{n}{2}}}} &=|x|\lim_{n\to\infty}\frac{1}{\sqrt{n}}\\ &=|x|\cdot{0}\\ &=0<1~\forall{x}\in\mathbb{R} \end{aligned}

4. I don't see how this is differnt than number three?

$\displaystyle \sum_{n=1}^{\infty}\frac{x^n}{n^{n^{\frac{1}{2}}}} =\sum_{n=1}^{\infty}\frac{x^n}{n^{\sqrt{n}}}$

Either note that $\displaystyle \color{red}\exists{N}\backepsilon\forall{n>N}\frac {x^n}{n^{\sqrt{n}}}\leqslant\frac{x^n}{n!}$

And since the right is the power series for $\displaystyle e$ which is convergent for all x the left side is as well (noting that the left side is strictly greater than zero)

Or just use the root test

\displaystyle \color{red}\begin{aligned}\lim_{n\to\infty}\sqrt[n]{\frac{x^n}{n^{\sqrt{n}}}}&=|x|\lim_{n\to\infty}\f rac{1}{n^{\frac{1}{\sqrt{n}}}}\\ &=|x|\cdot{0}\\ &=0<1~\forall{x}\in\mathbb{R} \end{aligned}

5. Using the Root test so we have that

\displaystyle \begin{aligned}\lim_{n\to\infty}\sqrt[n]{\frac{n^nx^n}{n!}}=&\lim_{n\to\infty}\frac{nx}{n! ^{\frac{1}{n}}}\\ &\sim\lim_{n\to\infty}\frac{nx}{(2\pi{n})^{\frac{1 }{2n}}ne^{-1}}\\ &=e|x|<1 \end{aligned}

I leave the endpoints up to you.

...
• Nov 17th 2008, 01:29 PM
Milus
Thank You very much for your help.
I calculated the endpoints and I would really appreciate if you just check them if I am not wrong.
1.set (-1,1) so no endpoints
2.the same as #1
3.all x
4.allx
5.[-1/e,1/e], so both endpoints belong to convergence set.
• Nov 17th 2008, 03:04 PM
Mathstud28
Quote:

Originally Posted by Milus
Thank You very much for your help.
I calculated the endpoints and I would really appreciate if you just check them if I am not wrong.
1.set (-1,1) so no endpoints(Yes)
2.the same as #1(Yes)
3.all x(Yes)
4.allx(Yes)
5.$\displaystyle \bigg[\frac{-1}{e},\frac{1}{e}{\color{red}\bigg)}$, so both endpoints belong to convergence set.(No)

...
• Nov 17th 2008, 04:30 PM
Milus
Thank You.
Why the point 1/e does not belong to convergence set?
isn't lim =0 if x =1/e and n goes to infinity?
• Nov 17th 2008, 05:14 PM
Mathstud28
Quote:

Originally Posted by Milus
Thank You.
Why the point 1/e does not belong to convergence set?
isn't lim =0 if x =1/e and n goes to infinity?

Consider that at the ponit $\displaystyle x=\frac{1}{e}$ we have that the summand is $\displaystyle \frac{n^n}{e^nn^n}\sim\frac{n^n}{e^n\sqrt{2\pi{n}} n^ne^{-n}}=\frac{1}{\sqrt{2\pi{n}}}$
• Nov 23rd 2008, 09:17 AM
drpawel
How to prove #5 using the fact that lim (n+(1/n))^n goes to e when n goes to infinity
• Nov 23rd 2008, 09:28 AM
Mathstud28
Quote:

Originally Posted by drpawel
How to prove #5 using the fact that lim (n+(1/n))^n goes to e when n goes to infinity

Is this a question or a statement?
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