
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
October 3rd, 2017, 09:13 AM  #1 
Newbie Joined: Oct 2017 From: UK Posts: 2 Thanks: 0  Pointwise, total convergence, and sum of a (power?) sequence
Hi guys... I seriously cannot help myself solving this problem: $\displaystyle \sum_1^\infty n^{2}e^{nx^2nx}$ (I'm not that good writing formulas but "n=1" and "+inf" are the sub and the sup) So I was asked to evaluate the pointwise and the total convergence of this sequence and find the sum. I have been thinking about to reduce it to a power sequence the fact is that I truly cannot think a way to do it.. Can you help me by resolving it and writing the steps? Thanks a lot! Best regards, An engineering student. 
October 3rd, 2017, 12:28 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,805 Thanks: 716 
I am not sure what you mean by evaluate. For x=0, divergent. Otherwise convergent. Power series doesn't exist (as far as I can tell).

October 3rd, 2017, 01:03 PM  #3 
Newbie Joined: Oct 2017 From: UK Posts: 2 Thanks: 0 
Yes, I probably make some mistakes while writing the first task of the exercise, ... maybe I got some problems with the translation since I am not English, sorry... but by the way I solved the first task.. what I cannot do now is to calculate the sum, that is the most difficult part...any ideas sir ?

October 5th, 2017, 01:34 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,805 Thanks: 716 
Good news! Sum is relatively easy. First note  divergence is for x in the closed interval from 1 to 0, not just 0. Sum: Recognize that $\displaystyle e^{x^2x}$ is an unnecessary complication. The problem is then $\displaystyle g(u)=\sum_{n=0}^\infty n^2u^n$, where $\displaystyle u=e^{x^2x}$. Starting at n=0 doesn't change anything. Step 1: $\displaystyle \sum_{n=0}^\infty n^2u^n=\sum_{n=0}^\infty n(n1)u^n+\sum_{n=0}^\infty nu^n=g_1(u)+g_2(u)$ Step 2: Look at $\displaystyle h(u)= \sum_{n=0}^\infty u^n=\frac{1}{1u}$, where u<1. Then $\displaystyle h'(u)=\sum_{n=0}^\infty nu^{n1}=\frac{1}{(1u)^2}$ and $\displaystyle h''(u)=\sum_{n=0}^\infty n(n1)u^{n2}=\frac{2}{(1u)^3}$ Put all this together to get $\displaystyle g_1=u^2h''(u)=\frac{2u^2}{(1u)^3}$ and $\displaystyle g_2=uh'(u)=\frac{u}{(1u)^2}$ so that $\displaystyle g(u)=\frac{u^2+u}{(1u)^3}$. 

Tags 
convergence, pointwise, power, sequence, sum, total 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Convergence of a Power Series  John Travolski  Calculus  3  November 29th, 2016 07:09 PM 
Sequence convergence (Leibniz and absolute convergence)  Chemist@  Calculus  7  November 1st, 2014 02:16 PM 
pointwise convergence  Sandra93  Real Analysis  1  May 14th, 2014 06:12 AM 
convergence of sum of prime numbers (with a power)  rkose  Number Theory  5  January 29th, 2010 07:08 AM 
Pointwise bounded sequence problem.  DinkyDoeDoe  Real Analysis  0  September 4th, 2009 12:33 PM 