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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.


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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
up vote 0 down vote accepted

In your assertion "The sum in the right bracket is less than one" you are basically avoiding all the work that needs to be done.

Since your sequence is a sequence of partial sums of a positive sequence, all the work reduces precisely to proving that the tails of the series are small.

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 $m>n>n_1$, $$ |x_{m+1}-x_n|<\varepsilon. $$

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@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
it all makes sense now... thanks – user44429 Nov 9 '12 at 3:43

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