Is the Infinite Sum of Continuous Functions Also Continuous?

JG89
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Homework Statement


Prove that the function f(x) = \sum_{n=1}^{\infty} \frac{cos(n^2x)}{e^{nx^2}2^n} is continuous on R.


Homework Equations





The Attempt at a Solution




I haven't learned about a series of functions converging to a function yet, but would it be sufficient to show that each function in the infinite sum is differentiable, so each function in the infinite sum is continuous, meaning that the sum of the functions must also be continuous?
 
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JG89 said:

Homework Statement


Prove that the function f(x) = \sum_{n=1}^{\infty} \frac{cos(n^2x)}{e^{nx^2}2^n} is continuous on R.


Homework Equations





The Attempt at a Solution




I haven't learned about a series of functions converging to a function yet, but would it be sufficient to show that each function in the infinite sum is differentiable, so each function in the infinite sum is continuous, meaning that the sum of the functions must also be continuous?

No, it is not enough that each summand be continuous for the sum to be continuous. What one usually does in such types of question is show that the sequence of partial sums is uniformly convergent. There is a variety of tests out there for that purpose. One of the most useful is the so-called Weierstrass M-test and sure enough, it can be used on the series that interests you: http://en.wikipedia.org/wiki/Weierstrass_M-test.
 
Thanks for the replies guys.

I haven't learned about uniform convergence yet, so I guess I don't have the proper "machinery" to attack this problem.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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