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How can one use the direct comparison test for $\displaystyle S=\sum_{k=1}^{\infty}{\frac{1}{\sqrt{k}+10k}}$?


CONTEXT: According to Wikipedia,

the comparison test, sometimes called the direct comparison test to distinguish it from similar related tests (especially the limit comparison test), provides a way of deducing the convergence or divergence of an infinite series or an improper integral. In both cases, the test works by comparing the given series or integral to one whose convergence properties are known.

MY THOUGHT: I think $S$ diverges since it looks similar to the harmonic series $\displaystyle \sum_{k=1}^{\infty}\frac{1}{k}$: if the term $\sqrt{k}$ is missing in $S$, then the series is obviously diverging. But I can't find a good series to directly compare the series $S$ to.

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3 Answers 3

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$\frac 1 {20k} \leq \frac 1 {\sqrt k +10k}$ and $\sum \frac 1 {20k}$ diverges.

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  • $\begingroup$ I think the downvotes are because of a lack of effort in all your answers $\endgroup$
    – Mathphile
    Commented Feb 23, 2020 at 2:34
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Let $a_k :=\frac{1}{\sqrt{k}+10k}$ and $b_k:=\frac{1}{\sqrt{k}}.$ Then show that $a_k/b_k \to 1$ as $k \to \infty.$ Hence there is $k_0$ such that $a_k/b_k \ge 1/2$ for $k> k_0.$

Can you proceed ?

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We have $$\sqrt{k}+10k< k+ 10k$$

Hence $$\frac1{\sqrt{k}+10k}>\frac1{11k}$$

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