Answered: "I am given the series∑n=1∞nn−1(2n2+n+1)(n+12)"

Lubna Boyce

Answered question

2022-02-28

I am given the series

\sum_{n = 1}^{\infty} \frac{n^{n - 1}}{{(2 n^{2} + n + 1)}^{(n + \frac{1}{2})}}

shotokan0758s

Beginner2022-03-01Added 8 answers

The ratio test works fine

a_{n} = \frac{n^{n - 1}}{{(2 n^{2} + n + 1)}^{(n + \frac{1}{2})}}

Take loragithms

\log (a_{n}) = (n - 1) \log (n) - (n + \frac{1}{2}) \log (2 n^{2} + n + 1)

Using Taylor for large values of n.

\log (a_{n}) = n \log (\frac{1}{2 n}) + \frac{1}{2} (\log (\frac{1}{2 n^{4}}) - 1) - \frac{5}{8 n} + O (\frac{1}{n^{2}})

Apply it twice and continue with Taylor series

\log (a_{n + 1}) - \log (a_{n}) = (\log (\frac{1}{2 n}) - 1) - \frac{5}{2 n}

\frac{a_{n + 1}}{a_{n}} = e^{\log (a_{n + 1}) - \log (a_{n})} = \frac{1}{2 e n} (1 - \frac{5}{2 e n} + O (\frac{1}{n^{2}}))

Now, using the expansion of

\log (a_{n})

O (\frac{1}{n})

, we then have your formula

a_{n} \sim \frac{1}{\sqrt{2 e}} \frac{1}{2^{n} n^{n + 2}}

which is an overestimate of the true

a_{n}

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