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I am trying to to show that the following series converges using the Comparison Test:

$\sum_{n=3}^\infty \frac{3\sqrt{n+1}}{\sqrt{n^4\log{n}+n\sin{n}+5}}$

So far, I have the following:

$0<\sum_{n=3}^\infty \frac{3\sqrt{n+1}}{\sqrt{n^4\log{n}+n\sin{n}+5}} < \sum_{n=3}^\infty \frac{3\sqrt{4n}}{\sqrt{n^4-n}} = \sum_{n=3}^\infty \frac{6}{\sqrt{n^3-1}}$

It is probably very obvious, but how does one deal with the $-1$ in the denominator to be able to get to the following?

$...<\sum_{n=3}^\infty \frac{k}{n^{3/2}}$

Thanks in advance!

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  • $\begingroup$ By the comparison test the series converges! $\endgroup$ – Dr. Sonnhard Graubner Apr 2 '18 at 19:09
  • $\begingroup$ If you're insistent on the comparison test, you can also use $\frac{6}{\sqrt{n^3-1}}<\frac{6}{\sqrt{(n-1)^3}}$ and shift the indices by one $\endgroup$ – Geoffroi Apr 2 '18 at 19:22
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$$n^3-1 > \frac{n^3}2$$

$$\frac1{\sqrt{n^3-1}}<\sqrt{\frac{2}{n^3}}$$

$$\sum_{n=3}^\infty\frac6{\sqrt{n^3-1}}<\sum_{n=3}^\infty\frac{6\sqrt2}{n^{3/2}}$$

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We have that:

$$\lim_{n \to \infty} \frac{\sqrt{n^3 - 1}}{n^{3/2}} = 1.$$

This implies that for $n \geq n_0$ we have $\sqrt{n^3 - 1} \geq \frac12 n^{3/2}$ and thus for $n \geq 3$ the fraction

$$ \frac{\sqrt{n^3 - 1}}{n^{3/2}} \geq \min\left\{ \frac{\sqrt{j^3 - 1}}{j^{3/2}}; j = 3, \ldots, n_0 \right\} \cup \{1/2\} = c > 0$$

since it is a minimum of finitely many terms.

The argument above is the core of the comparison test; if $a_n \sim b_n$ as $n \to \infty$ then for $n \geq n_0$ we have

$$ \frac12 \ b_n \le a_n \le \frac32\ b_n \ \text{for all $n \geq n_0$}, $$

implying

$$ \frac12\ \sum_{n \geq n_0} b_n \le \sum_{n \geq n_0} a_n \le \frac32\ \sum_{n \geq n_0} b_n. $$

Recall that the convergence of a series does not depend on its first terms.

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The limit comparison test can save you many headaches. You should not try to worry about bounding the small stuff, it should simply be ingored.

Factor out the "big parts" from the numerator and the denominator. This results in $$ \sum_{n=3}^\infty \frac{\sqrt{n}}{n^2\log n}\cdot\left(\frac{3\sqrt{1+1/n}}{\sqrt{1+\frac{\sin{n}}{n^3 \log n}+\frac{5}{n^4 \log n}}}\right) $$ Notice that the huge quotient on the right converges to $1$ as $n\to\infty$. By the limit comparison test, your series converges if and only if $$ \sum_{n=3}^\infty \frac{\sqrt{n}}{n^2\sqrt{\log n}}=\sum_{n=3}^\infty \frac{1}{n^{3/2}\sqrt{\log n}} $$ does. This is now easy to apply the usual comparison test to.

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Note that

$$\frac{3\sqrt{n+1}}{\sqrt{n^4\log{n}+n\sin{n}+5}}\sim \frac{3}{n^{\frac32}\log^{\frac12} n}$$

then the given series converges by limit comparison test with $\sum \frac{3}{n^{\frac32}\log^{\frac12} n}$ which indeed converges by limit comparison test with $\sum \frac{1}{n^{\frac54}}$.

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