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determine whether the following infinite series converges or diverges, please state what convergence/divergence test was used.

$$\sum_{k=1}^{\infty} \frac{4^k+5}{5^k + k}$$

I have some clue on solving this one. I think it converges because I know that that $\sum_{k=1}^{\infty} (\frac{4}{5})^k$ converges, and adding $k$ at the bottom instead of a fixed $5$ at the top will make the series smaller than $(\frac{4}{5})^k$, but is there a more algebraic/formal way of proving this?

How can I arrive algebraically at what the series would converges to?

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    $\begingroup$ There is no reason to believe this converges to anything "nice." You can prove convergence by using a suitable Comparison. We do need to worry (but not much) about the $5$ in the numerator. Or use Limit Comparison. $\endgroup$ – André Nicolas Oct 14 '13 at 17:46
  • $\begingroup$ Oh I believe it converges! Isn't (4/5)^k a suitable comparison? $\endgroup$ – Ted Flethuseo Oct 14 '13 at 17:51
  • $\begingroup$ @TedFlethuseo: You are correct, but you need not expect a nice answer for your very last question. You simply know it converges (but not what it actually converges to). $\endgroup$ – anon271828 Oct 14 '13 at 18:09
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    $\begingroup$ T0 $100$ digits precise, it converges to $4.871534897449792396910736095040374329744953119548855313188201330501084394593572900627247672463071002$, if that sparks someone's mind. WA could not find an exact match $\endgroup$ – Jean-Sébastien Oct 14 '13 at 18:26
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Comparison test (for $k>1$): $$0<\frac{4^k+5}{5^k+k}\le \frac{4^k+4^k}{5^k}=2\left(\frac{4}{5}\right)^k,$$

since $\sum 2\left(\frac{4}{5}\right)^k$ is converging, the dominated series must converge too.

Alternatively, by the Limit Comparison test: $$\lim_{k\to\infty}\frac{\frac{4^k+5}{5^k+k}}{\frac{4^k}{5^k}}=\lim_{k\to\infty}\frac{1+\frac{5}{4^k}}{1+\frac{k}{5^k}}=1,$$ therefore two series $\sum\frac{4^k+5}{5^k+k}$ and $\sum \frac{4^k}{5^k} $ are equivalent in their convergence (i.e. both are converging or diverging). The last one is converging geometric, therefore the first one is converging as well.

Where does it converge to? What a great question!

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  • $\begingroup$ How did you get to (1+5/4^k)/(1+k/5^k)? $\endgroup$ – Ted Flethuseo Oct 14 '13 at 18:48
  • $\begingroup$ @TedFlethuseo I divided corresponding numerators and denominators using the fact that $(a+b)/c=a/c+b/c$ $\endgroup$ – AstroSharp Oct 14 '13 at 19:40
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For every $k\geqslant1$, $$ \frac{4^k+5}{5^k+k}\leqslant\frac{4^k+5}{5^k}=a^k+5\cdot b^k,\qquad a=\frac45,\qquad b=\frac15. $$ Since $a\lt1$ and $b\lt1$, this shows that the series converges.

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  • $\begingroup$ Nitpick: you also need a lower bound. Trivially, zero will do. $\endgroup$ – Adrian Ratnapala Oct 14 '13 at 18:34
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    $\begingroup$ @AdrianRatnapala Trivially. $\endgroup$ – Did Oct 14 '13 at 18:34
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    $\begingroup$ Mea Culpa, the lower bound is irrelevant - a sequence could wonder up and down between two bounds. The real point is that a sequence is monotonically approaching some bound which it cannot do without a limit. $\endgroup$ – Adrian Ratnapala Oct 14 '13 at 18:38
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Note that your premise is incorrect, and the given summand is in fact greater than $\left(\frac{4}{5}\right)^k$ for all $k>N$ for some $N$ (which turns out to be at least $1$, but this is irrelevant for convergence). If$$\frac{4^k+5}{5^k+k}>\frac{4^k}{5^k}$$ is true then $$20^k+5^{k+1}>20^k+k4^k$$ $$\left(\frac{5}{4}\right)^k>\frac{k}{5}.$$ Note that this is satisfied when $k=100$, as $(1+\frac{1}{4})^{100}\ge1+\frac{100}{4}>\frac{100}{4}$.

Both left and right sides are increasing functions. Comparing derivatives,$$\left(\left(\frac{5}{4}\right)^k\right)'=\left(\frac{5}{4}\right)^k\ln\left(\frac{5}{4}\right)>\left(\frac{k}{5}\right)'=\frac{1}{5}$$ and we are done (the last inequality is obviously satisfied for all $k>N$ for some $N$, as $\left(\frac{5}{4}\right)^k\ln\left(\frac{5}{4}\right)$ is increasing whereas $\frac{1}{5}$ is not).

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A cleaner hint is to notice that if $k\geqslant1$, then $$\frac{1}{5^k}\geqslant\frac{1}{5^k+k}\geqslant\frac{1}{6^k}$$

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