# Proof that the sequence $x_{n} = 1 + \frac{1^2}{7^1} + \frac{2^2}{7^2} + … + \frac{n^2}{7^n}$ is Cauchy.

Since I'm new to analysis, I'm still never sure if my proofs are sound or have logical holes in them. Here's my proof for this one - hopefully you guys can find whatever logical errors are in the proof, or suggest a more efficient, slicker proof.

So, to show that a sequence is Cauchy, we look at the distance between consecutive terms further down the sequence. If we consider $x_{n}$ and $x_{m}$, then we see that if $m > n$ we can ignore all the terms in the difference of these two sums up to and including $\frac{n^2}{7^n}$ and consider the distance between $\frac{n^2}{7^n}$ and $\frac{(n+1)^2}{7^{n+1}}$, that is $\left|{x_{n} - x_{n+1}} \right|$, the distance $\left|{x_{n+1} - x_{n+2}} \right|$ and so on up to $\left|{x_{m-1} - x_{m}} \right|$. Now I'm too lazy to write it all down here, but we see that the distance between any two terms $\left|{x_{n} - x_{n+1}} \right|$ of this difference of sums is smaller than $(\frac{6}{7})(\frac{n^2}{7^n})$. So considering the sum of all these distances, we see that this is simply $(\frac{6}{7})(\frac{n^2}{7^{n}} + \frac{(n+1)^2}{7^{n+1}} +...+\frac{(m-1)^2}{7^{m-1}})$. The sum in the right bracket is less than one (or is it? why?...) so that whole expression is less than 6. Given $\epsilon > 0$, and with $N = 6$, $m, n > N$, and $m \geq n$ we have that $\left|{x_{m} - x_{n}} \right| < \epsilon$, which shows the sequence is Cauchy.

Shoot..

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You cannot choose $\epsilon$: it's given to you as a challenge and you have to produce $N$ such that $|x_m-x_n|<\epsilon$ for all $m,n>N$. – lhf Nov 7 '12 at 23:40
sorry yeah i fixed some stuff – user44429 Nov 7 '12 at 23:42
If you know about convergence tests for series, you can use the ratio test to prove that the series converges, which implies that it is Cauchy. – lhf Nov 7 '12 at 23:51
no ratio test yet.. – user44429 Nov 7 '12 at 23:52
I thought so... Working out this problem will give you a good appreciation for the tests latter. – lhf Nov 7 '12 at 23:53

Here, what you can do is the following: for $n>n_0$ for some suitable $n_0$, $n^2<6^n$. So, for $n,m<n_0$, $$x_{m+1}-x_n=\sum_{k=n}^m\frac{k^2}{7^k}<\sum_{k=n}^m\frac{6^k}{7^k}=\sum_{k=n}^m\left(\frac{6}{7}\right)^k =\frac{(6/7)^n-(6/7)^{m+1}}{1-6/7}<\frac{(6/7)^n}{1-6/7}=\frac{6^n}{7^{n-1}}$$ Now if you fix $\varepsilon>0$, then you can choose $n_1>n_0$ and such that $6^n/7^{n-1}<\varepsilon$. Then, for any $m>n>n_1$, $$|x_{m+1}-x_n|<\varepsilon.$$
@pootieman: He’s using the fact that $\sum_{k\ge0}\frac1{2^k}=2$, so $$\frac1{2^{n-1}}\sum_{k=0}^{m-n-1}\frac1{2^k}<\frac1{2^{n-1}}\cdot2=\frac1{2^{n‌​-2}}\;,$$ where the inequality follows from the fact that the missing terms are positive. – Brian M. Scott Nov 8 '12 at 0:21
so when you chose n large such that it satisfies $n^2 < 6^n$, was that completely arbitrary? could you have chosen any base (> 1) for the exponential? is using an exponential in n in the numerator the obvious way to prove this sequence is cauchy or is there a kind of "slicker" method? thanks again. – user44429 Nov 9 '12 at 1:53
One of the most common tricks in the book is to use geometric series, because they are among the very few that can be explicitly calculated. The choice of $6$ is to get the fraction $6/7<1$; as you say, any number in the interval $(1,7)$ would have done (bigger than $1$ so that it eventually wins over $n^2$, less than $7$ to get a fraction $<1$). As for you second question, I think this is the canonical way (which can be summarized in the "ratio test"). – Martin Argerami Nov 9 '12 at 2:31