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May 19, 2013

# Homework Help: Calculus2

Posted by Ana on Thursday, April 21, 2011 at 3:19am.

Please, help me I really need help on this problem. Thank you so much for all your help.
Question: Find the radius of convergence and interval of convergence for
sum(n=1to infinity)(-1)^n-1 n(x-1)^n/(n^2+1)2^n

Thank you again

• Calculus2 - MathMate, Thursday, April 21, 2011 at 11:44am

The summation is for n=1 to ∞.
The expression is an alternating series:

(-1)^(n-1) * n(x-1)^n / ((n²+1)2^n)

You would be using different tests for convergence and choose values of x between which the series is convergent.

For example, the limit test requires that an=0 as n->∞, so
For an=0 at n->∞, we require
Lim n(x-1)^n / ((n²+1)2^n) = 0
which is equivalent to requiring convergence of ((x-1)/2)^n, which in turn implies |x-1|<2, or x0±r, where x0=1, r=2.

Note that the at the extreme limit of radius of convergence (-1 and 3 in this case), convergence has to be checked individually.
At x=3, we are looking at the limit of an alternating series:
(-1)^(n-1) n(x-1)^n / ((n²+1)2^n)
which reduces at x=3 to
(-1)^(n-1) n/(n^sup2;+1)
or
(-1)^(n-1) 1/n
which converges by comparison with Leibniz series of
1-1/2+1/3-1/4+1/5-...

Similarly, the series converges at x=-1.

So the radius of convergence is 0≤r≤2.

Similar tests can and should be applied:
- ratio test
requires an+1/an <1.
- root test
requires (an)^(1/n)<1
- leibniz test for alternating series
basically requires both the ratio test and limit test to pass.

Check out all the tests available to you, from above and from your class notes, and make a final decision on the radius of convergence.

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