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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.

Shoot..

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    I thought so... Working out this problem will give you a good appreciation for the tests latter.2012-11-07

1 Answers 1

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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, $ 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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    it all makes sense now... thanks2012-11-09