How could I show that the series $$\sum_{n=1}^\infty\frac1{\sqrt{n^2+1}}$$ Diverges without the use of comparison test which I am able to show. Any hints would be great thanks.
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Thanks Parcly – Eden Hazard May 21 '19 at 16:27
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2What’s wrong with the comparison test? – mheldman May 21 '19 at 16:29
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I meant without using comparison test , added “limit” by accident – Eden Hazard May 21 '19 at 16:29
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1You could use the integral test! – Angina Seng May 21 '19 at 16:51
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Hint:With the integral test, we have that $$\int_0^{\infty} \frac{1}{\sqrt{x^2+1}}\mathrm{d}x$$ and $$\sum_{n=0}^{\infty} \frac{1}{\sqrt{n^2+1}}$$ are equiconvergent.
Botond
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I have never read the word equiconvergent but I agree with your answer! – manooooh May 21 '19 at 16:48
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1@manooooh I think it's not a common thing, but I saw it in an analysis note and I started to like it :) – Botond May 21 '19 at 16:55
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You can use the integral test:$$\int_0^\infty\frac{\mathrm dx}{\sqrt{x^2+1}}=\lim_{M\to\infty}\log\left(M+\sqrt{1+M^2}\right)=\infty.$$
José Carlos Santos
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To avoid the comparison test, compute $$\lim_{n\to \infty} \frac{n}{1}\frac{1}{\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{1}{1+n^{-2}}=1$$
Hence, by the limit comparison test, the series agrees with $$\sum_{n\ge 1}\frac{1}{n}$$ which diverges.
vadim123
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without the use of comparison test which I am able to show – I was suspended for talking May 21 '19 at 16:31
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