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Prove the following sequence is convergent: $$ x_n = 1 + {1^2\over 4} + {2^2\over 4^2} + \cdots + {n^2\over 4^n} $$

I've decided to try Cauchy Criterion here. Consider $|x_m - x_n|$ for $m = 2n > n$: $$ \begin{align} |x_m - x_n| &= \left| 1+ \sum_{k=1}^m\frac{k^2}{4^k} - 1 - \sum_{k=1}^n\frac{k^2}{4^k}\right|\\ &= \left|\sum_{k=n+1}^{2n}\frac{k^2}{4^k}\right| \\ &\stackrel{>0}{=} \sum_{k=n+1}^{2n}\frac{k^2}{4^k} \\ &\le \sum_{k=n+1}^{2n}\frac{(2n)^2}{4^k} \\ &\le \sum_{k=n+1}^{2n}\frac{(2n)^2}{4^{n+1}} \\ &= \frac{4n^3}{4^{n+1}} \\ &= \frac{n^3}{4^n} \end{align} $$

Consider the limit: $$ 0 \le \lim_{n\to\infty}|x_{2n} - x_n| \le \lim_{n\to\infty} \frac{n^3}{4^n} = 0 $$

Which by squeeze theorem gives: $$ \lim_{n\to\infty}|x_{2n} - x_n| = 0 $$

proving $x_n$ is a fundamental sequence, hence convergent.

I've been also thinking of using monotone convergence theorem: $$ x_{n+1} - x_n = \frac{(n+1)^2}{4^{n+1}} > 0 $$ By this $x_n$ is monotonically increasing, but i couldn't find an upper bound.

I would like to know two things:

  1. It the above a rigorous way to show $x_n$ is convergent?
  2. Even though the problem statement only asks to prove convergence rather than find the limit I would still want to find it. W|A suggests its ${20\over 27}$. Is it possible to obtain that result using more or less elementary calculus reasoning?(anything before the definition of a derivative)
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Hint

For $x>4$ we have $$2^x>x^2$$therefore for $n>4$ we obtain $${n^2\over 4^n}<{2^n\over 4^n}={1\over 2^n}$$

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Using ratio test: $$\lim_{n\to\infty}|\frac{(n+1)^2 4^{-(n+1)}}{n^2 4^{-n}}|=\frac{1}{4}<1$$ so the sequence of partial sums is convergent. Furthermore, by considering the geometric series $$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$ $$\frac{1}{(1-x)^2}=\sum_{n=0}^\infty nx^{n-1}$$ $$f(x)=\frac{x}{(1-x)^2}=\sum_{n=0}^\infty nx^n$$

$$\frac{f'(\frac{1}{4})}{4}=\sum_{n=0}^\infty \frac{n^2}{4^{n}}$$ you can easily deduce the limit of the sequence.

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    $\begingroup$ It is just adding a constant and does not depend on $n$, so it should not affect convergence. $\endgroup$ – aleden Jan 9 at 16:13
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$\frac{n^2}{4^n}=\frac{1}{2^n}\frac{n^2}{2^n}\leq C\frac{1}{2^n}$ for some constant $C,$ because $\frac{n^2}{2^n}\rightarrow 0$ and hence bounded sequence. So, $\sum_n \frac{n^2}{4^n}\leq C\frac{1}{2^n}<\infty.$

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