Otdfbt

Let $(x_n)$ be a decreasing sequence with $x_n > 0$ for all $n in mathbbN$, and $(x_n) to 0$. Let $(y_n)$ be defined for all $n in mathbbN$ by
$$y_n = x_0 - x_1 + x_2 - cdots + (-1)^n x_n .$$

I want to show, using the $varepsilon$ definition, that $(y_n)$ is Cauchy.

I am trying to find, given $varepsilon > 0$, a real number $N$ such that for all $m$ and $n$ with $m > n > N$, $|y_m - y_n| < varepsilon$.

I have been going backwards to try and find $N$, and have
beginalign*
|y_m - y_n| & = left| (x_0 - x_1 + cdots pm x_m) - (x_0 - x_1 + cdots pm x_n) right| \
|y_m - y_n| & = left| x_n + 1 - x_n + 2 + cdots pm x_m right| \
|y_m - y_n| & leq | x_n + 1 | + | x_n + 2 | + cdots + | x_m | \
|y_m - y_n| & leq ?
endalign*

I do not know how to get a solution from there, and am not sure about the process, particurlary the last step since I feel getting rid of the minuses might prevent me from finding a solution.

To see that the sequence of partial sums is Cauchy, you cannot use the triangle inequality directly as you did. A famous counter example here is $sum_k=1^inftyfrac(-1)^kk$.

What you can do is grouping the terms of the partial sums $s_n= sum_j=1^n(-1)^jx_j$ as follows:

• Let $m = n+k, k,n in mathbbN$
  

Now, you can write $|s_m - s_n|$ in two different ways:

$$|s_n+k - s_n| = begincases
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-x_n+2i+1)| & k = 2i+1 \
|x_n+1 - (x_n+2-x_n+3) - cdots - (x_n+2i-2-x_n+2i-1) - x_2i| & k = 2i \
endcases
$$

$$|s_n+k - s_n| = begincases
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) + x_n+2i+1| & k = 2i+1 \
|(x_n+1 - x_n+2) + cdots + (x_n+2i-1-x_n+2i) | & k = 2i \
endcases
$$

Using the fact that $x_n searrow 0$, it follows immediately that for all $k in mathbbN$ holds
$$|s_n+k - s_n| leq x_n+1$$

Hence, for $epsilon > 0$ choose $N_epsilon$ such that $x_N_epsilon < epsilon$. Then, for all $m> n > N_epsilon$ you have
$$|s_m - s_n| leq x_n+1 leq x_N_epsilon < epsilon$$

This is also known as the "Leibnitz's Test".

We write $s_n = x_1-x_2+x_3-...+(-1)^n+1x_n$

$s_2n+2-s_2n=u_2n+1-u_2n+2 geq0$ for all $n$.

$s_2n+1-s_2n-1=-u_2n+u_2n+1 leq 0$

$s_2n =u_1 -(u_2-u_3)-(u_4-u_5)...-u_2n leq u_1$, i.e. a monotone increasing sequence bounded above.

$s_2n+1 =(u_1 -u_2)+(u_3-u_4)+...+u_2n+1 geq u_1-u_2$, i.e. a monotone decreasing sequence bounded below.

Hence, both are convergent subsequences of $(s_n)$. But, we have $lim (s_2n+1-s_2n)=u_2n+1=0$, therefore, they converge to the same limit.

Hence, $(s_n)$ converges, i.e. it is Cauchy.

Note: We conclude that $(s_n)$ converges because the indices of the two subsequences $(s_2n)$ and $(s_2n+1)$ i.e. $U = 2n+1 : n in mathbbN$ and $V = 2n : n in mathbbN$ form a partition of $mathbbN$ and they both converge to the same limit.