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Consider the series $ ∑_{n=1}^∞ x^2+ n/n^2$ . Pick out the true statements:

Consider the series $ \sum_{n=1}^\infty (-1)^n(x^2+ n)/n^2$
Pick out the true statements:
(a) The series converges for all real values of $x$.
(b) The series converges uniformly on $\mathbb{R}$.

how can I able to solve this.

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marked as duplicate by MJD, Ittay Weiss, Austin Mohr, Micah, Brandon Carter Jan 14 '13 at 6:19

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Use ideas in a very similar question posted minutes earlier. –  Calvin Lin Jan 14 '13 at 3:09
    
The terms $(-1)^n{x^2+n\over n^2}$ do not converge uniformly to $0$ on $\Bbb R$. Thus, the series can't converge uniformly on $\Bbb R$ (a series converges uniformly if and only if it is uniformly Cauchy). –  David Mitra Jan 14 '13 at 3:33
    
Incidentally, an "alternating series" converges uniformly if the terms converge uniformly to $0$. –  David Mitra Jan 14 '13 at 3:42
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2 Answers 2

Splitting up the infinite series:

$$\sum_{n=1}^{\infty} (-1)^{n} (x^2 + n)/n^{2} = \sum_{n=1}^{\infty} (-1)^{n} x^{2}/n^{2} + \sum_{n=1}^{\infty} (-1)^{n} n/n^{2} = x^2 \sum_{n=1}^{\infty} (-1)^{n}/n^{2} + \sum_{n=1}^{\infty} (-1)^{n}/n$$

Note that each of the last two infinite sums converges (why?) which justifies the above line read from right to left. In particular, the sum of two convergent series converges.

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You can apply alternating series test

http://en.wikipedia.org/wiki/Alternating_series_test

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I got (a).but not sure about (b) –  ketu Jan 14 '13 at 3:27
    
you can use David Mitra arguments above on the comments –  user52188 Jan 14 '13 at 4:13
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