Inequality: $\left|\sum_{i = n + 1}^{\infty} \frac{(-1)^i}{i + x}\right| \le \frac{1}{n + 1 + x}$ This (with $x > 0$ and $n \in \Bbb N$):
$$ \left|\sum_{i = n + 1}^{\infty} \frac{(-1)^i}{i + x}\right| \le \frac{1}{n + 1 + x}$$
Was used to prove that the series of general term the summand on the left hand side of the inequality is convergent uniformly on $]0, \infty[$. But I can't see why this should be true. 
Using the $|\sum \cdot| \le \sum|\cdot|$ does nothing, because we get an upper bound which is greater than the RHS. It seems to be true after observing the first few terms of the sum. But how could it be proven? There's just some trick which I'm not seeing. 
Thanks a lot.
 A: Hint 1: The terms $$ \frac{(-1)^i}{i + x} $$ 
are of alternating signs and their absolute value decreases.
Hint 2: Group the terms in the sum into pairs of two.
More detailed answer: 
$$
\left| \sum_{i = n + 1}^{\infty} \frac{(-1)^i}{i + x}\right|
= | (-1)^{n+1} | \left| \Bigl( \frac{1}{n + 1 + x} - \frac{1}{n + 2 + x} \Bigr)
+ \Bigl( \frac{1}{n + 3 + x} - \frac{1}{n + 4 + x} \Bigr)
+ \dots \right|
$$
All terms in the parentheses are positive, so this is equal to 
$$
 \Bigl( \frac{1}{n + 1 + x} - \frac{1}{n + 2 + x} \Bigr)
+ \Bigl( \frac{1}{n + 3 + x} - \frac{1}{n + 4 + x} \Bigr)
+ \dots 
$$
Now we group the sum differently:
$$
\frac{1}{n + 1 + x} + \Bigl( - \frac{1}{n + 2 + x} + \frac{1}{n + 3 + x} \Bigr)
+ \Bigl( - \frac{1}{n + 4 + x} + \frac{1}{n + 5 + x} \Bigr)
+ \dots
$$
All terms in the parentheses are now negative, and the result follows.
(A more rigorous proof would work with finite partial sums, but I hope this
demonstrates the idea.)
A: If you group pairs of terms together you get 
$$ \frac{(-1)^i}{i + x} - \frac{(-1)^i}{i + 1 + x}  = \frac{(-1)^i}{(i + 1 + x) (i+x)}$$
So you can bound the sum by:
$$|\sum_{i = n + 1}^{\infty} \frac{(-1)^i}{i + x}| \le \sum_{\substack{i = n + 1 \\ (i-n) \textrm{ is odd}}}^{\infty} \frac{1}{(i + 1 + x) (i+x)} \le \sum_{i = n + 1}^{\infty} \frac{1}{(i + 1 + x) (i+x)}$$
Since $\int_{i}^{i+1} \frac{1}{(t+x)^2} dt = \frac{1}{x+i} - \frac{1}{x+i+1} = \frac{1}{(x+k)(x+k+1)}$, you can compute the above series explicitly as $$\sum_{i = n + 1}^{\infty} \int_{i}^{i+1} \frac{1}{(t+x)^2} dt =\int_{n+1}^\infty \frac{1}{(t+x)^2} dt = \frac{1}{n+1+x}$$

Alternative way
An alternative way to compute the sum is to break it up again:
$$\sum_{i = n + 1}^{\infty} \frac{1}{(i + 1 + x) (i+x)} =  \sum_{i = n + 1}^{\infty} \left( \frac{1}{i+x} - \frac{1}{i + 1 + x} \right)$$
and notice that this is a telescoping series so the negative term for $i$ cancels with the positive term for $i+1$, and we are left with only the positive term for $i=n+1$, i.e. $\frac{1}{n+1+x}$
