Show that $\displaystyle \sum_{j=1}^{\infty} \frac{-2j}{(x^2 + j^2)^2}$ converges uniformly.
Don't know how to do this problem since $x$ and $j$ are in the expression together. Is there a convergence test I can use?
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1Let $s_n$ denote the sequence of partial sums. Then, the sum converges uniformly iff $s_n(x)$ converges uniformly as $n\to\infty$. – Apr 19 '14 at 18:26
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Ok, but is there a convergence test I can use for the expression inside the sum? – kiwifruit Apr 19 '14 at 18:26
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Have you heard of Abel's Uniform Convergence Test? – Apr 19 '14 at 18:27
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No, unfortunately that's not something that was covered – kiwifruit Apr 19 '14 at 18:29
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See Mathworld Wolfram on this. – Apr 19 '14 at 18:29
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Thank you, I will read about this, but is there another method, since I am not able to use this in class? – kiwifruit Apr 19 '14 at 18:30
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I don't remember... Sorry – Apr 19 '14 at 18:31
2 Answers
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$$\sum_{j=1}^\infty\frac{-2j}{(x^2+j^2)^2}\le\sum_{j=1}^\infty\frac{-2j}{j^4}= -2\sum_{j=1}^\infty\frac{1}{j^3}$$
Hence it can be majorized by the convergent series $-2\sum_{j=1}^\infty\frac{1}{j^3}.$
Mher
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*Hint: you can use the M-test
$$\frac{2j}{(x^2 + j^2)^2} \leq \frac{2j}{( j^2)^2}.$$
Mhenni Benghorbal
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