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Determine the convergence or divergence of f(n) = nan

Consider the function f(n) defined by

    \[ f(n) = na^n \qquad \text{where} \quad |a| < 1. \]

Determine whether the sequence \{ f(n) \} converges or diverges, and if it converges find the limit.

We use the definition of the exponential and some tricks to get f(n) into a form we know how to deal with,

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

The final line follows since

    \[ \lim_{n \to \infty} \left(\log a + \frac{\log n}{n}\right) < 0 \]


    \[ \lim_{n \to \infty} \frac{\log n}{n} = 0 \qquad \text{and} \qquad |a| < 1. \]


    \[ n \left( \log a + \frac{\log n}{n} \right) \to -\infty \qquad \text{as} \qquad n \to \infty.\]

Thus, \{ f(n) \} converges to the limit 0.


  1. alberto balsalm says:

    Would it still be valid to express the question as

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    These links might help in finding solution:

    and just claim the limits of both the numerator and denominator approach 0?

    • Mohammad Azad says:

      And? it’s an indeterminate form so your only hope is L’Hopital’s rule which makes it worse.
      Here is my solution: if a is zero the limit is trivial, so suppose a>0, then the limit becomes n/e^(-nloga) and since abs(a)0 hence by theorem 7.11 the limit is zero. Now if a0 so the limit of n(-a)^n is zero by the argument above so for every epsilon>0 there is a delta>0 such that if n>delta then abs(n(-a)^n)<epsilon this proves this case since abs(n(a)^n)=abs(n(-a)^n)

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