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$\displaystyle\sum_{k=1}^\infty (2k)!/k!(k+1)!$

Let $a_k = (2k)!/k!(k+1)!$

$\lvert a_{k+1}/a_k\rvert \to 4$ as $k \to \infty$

Thus the series is divergent. Can someone double check ... my gut says it is convergent.

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  • $\begingroup$ See also this if you want to apply the root test. $\endgroup$ – lhf Jul 11 '14 at 12:47
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Alternate.

Binomial coefficient $$ \frac{(2k)!}{(k!)^2} $$ is a positive integer, thus $\ge 1$, so your series is $\ge$ $$ \sum_{k=1}^\infty\frac{1}{k+1} $$ which diverges.

Of course, the middle binomial coefficient is much bigger than $1$, so divergence is much worse than the harmonic series.

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  • $\begingroup$ Some simple bounds for the central binomial coefficient appear in Wikipedia. $\endgroup$ – lhf Jul 11 '14 at 12:43
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conducting the Ratio test:

You series is $$\frac{2k!}{k!(k+1)!}$$

Now, $$\lim_{k\rightarrow \infty} \frac{\frac{(2k+2)!}{(k+1)!(k+2)!}}{\frac{2k!}{k!(k+1)!}}$$

$$\lim_{k\rightarrow \infty} \frac{(2k+2)!}{(k+1)!(k+2)!}*\frac{k!(k+1)!}{2k!}$$ $$\lim_{k\rightarrow \infty} \frac{k!(2k+2)!}{(2k)!(k+2)!}$$ $$\lim_{k\rightarrow \infty} \frac{(2k+1)*(2k+2)}{(k+1)*(k+2)}$$ $$\lim_{k\rightarrow \infty} \frac{(2k+1)*(2k+2)}{(k+1)*(k+2)} = 4$$

You are correct.

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