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Test the given series for convergence or divergence

Determine if the following series converges or diverges and justify your decision.

    \[ \sum_{n=1}^{\infty} \frac{n^{n+\frac{1}{n}}}{\left( n + \frac{1}{n} \right)^n}. \]

The series diverges. We will show this by showing that the terms a_n do not go to 0. To that end,

    \begin{align*}  \lim_{n \to \infty} a_n &= \lim_{n \to \infty} \frac{n^{n + \frac{1}{n}}}{\left( n + \frac{1}{n} \right)^n} \\[9pt]  &= \lim_{n \to \infty} \frac{e^{\left( n + \frac{1}{n} \right) \log n}}{e^{n \log \left( n + \frac{1}{n} \right)}} \\[9pt]   &= \exp \left( \lim_{n \to \infty} \left( \left( n + \frac{1}{n} \right) \log n - n \log \left( n + \frac{1}{n} \right) \right) \right) \\[9pt]  &= \exp \left( \lim_{n \to \infty} \left( n (\log n - \log(n+1/n)) + \frac{\log n}{n} \right) \right). \end{align*}

So, we’d like to split this limit of a sum into a sum of limits, which we can do if the each of the limits exist on their own. We know \lim_{n \to \infty} \frac{\log n}{n} = 0, so we need to worry about the first term. We compute its value (and the fact that it’s finite will justify splitting the limit into two separate limits),

    \begin{align*}  \lim_{n \to \infty} \left( n (\log n - \log (n + 1/n)) \right) &= \lim_{n \to \infty} \left( n \log \left( \frac{n}{n+\frac{1}{n}} \right) \right) \\[9pt]  &= \lim_{n \to \infty} \frac{\log \left( \frac{1}{1+\frac{1}{n^2}} \right)}{\frac{1}{n}}. \end{align*}

Now, we can apply L’Hopital’s rule since this is the indeterminate form 0/0,

    \begin{align*}  &= \lim_{n \to \infty} \left( -\frac{1}{n^2} \right)^{-1} \cdot \frac{2}{n^3+n} \right) \\[9pt]  &= \lim_{n \to \infty} \frac{-2n}{n^2+1} \\[9pt]  &= 0. \end{align*}

So, putting this all back together we have,

    \begin{align*}  \lim_{n \to \infty} a_n &= \exp \left( \lim_{n \to \infty} \left( n (\log n - \log(n+1/n)) + \frac{\log n}{n} \right) \right) \\[9pt]  &= \exp \left(\lim_{n \to \infty} \left(n (\log n - \log (n+1/n))\right) + \lim_{n \to \infty} \frac{\log n}{n} \right) \\[9pt]  &= e^{0 + 0} \\[9pt]  &= 1. \end{align*}

Hence, the terms in the series do not tend to 0, so the series cannot converge.


  1. Mohammad Azad says:
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    \[\lim_{n\to\infty}\frac {n^{n+1/n}}{(n+1/n)^n}=\lim_{n\to\infty}\left(\frac {n}{n+1/n}\right)^nn^{1/n}\ge \lim_{n\to\infty}\left(\frac {n}{n+1}\right)^nn^{1/n}=\frac {1}{e}>0\]
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  2. Mohammad Azad says:

    \lim_{n\to\infty}\frac {n^{n+1/n}}{(n+1/n)^n}=\lim_{n\to\infty}\left(\frac {n}{n+1/n}\right)^nn^{1/n}≥\lim_{n\to\infty}\left(\frac {n}{n+1}\right)^nn^{1/n}=\frac {1}{e}>0

  3. Artem says:

    A simpler way would be to find a different lower bound which would diverge. Here the bound would be: \frac{n^n}{(n + \frac{1}{n})^n}. Then the limit of its term sequence shows they tend to 1. The limit can be computed with the Binomial theorem decomposition.

    • Mohammad Azad says:

      \lim_{n\to\infty}\frac {n^{n+1/n}}{(n+1/n)^n}=\lim_{n\to\infty}\left(\frac {n}{n+1/n}\right)^nn^{1/n}≥\lim_{n\to\infty}\left(\frac {n}{n+1}\right)^nn^{1/n}=\frac {1}{e}>0

  4. tom says:

    I like how you used L’Hopital’s rule here but a simpler method may suffice: n^(n+1/n)/(n+1/n)^n < n^(n+1/n)/n^n = n^(1/n), which does not go to zero.

    • Mohammad Azad says:

      That doesn’t imply anything, since being less than a term which goes to one doesn’t mean that the terms don’t go to zero

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