Test the following series for convergence..

Jan 2010
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\(\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^{sinh(n)}}\)
 

chisigma

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Because is \(\displaystyle \sinh 1 = 1.1752011936438...\) and the function \(\displaystyle \sinh (*)\) is [strongly] increasing, for all n will be...

\(\displaystyle \frac{1}{n^{\sinh n}} < \frac{1}{n^{1.1}}\) (1)

Now the series...

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^{1.1}}\) (2)

... converges so that...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)