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I want to answer whether the following series is convergent or divergent:

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n+\frac{10100}{n}}$$

Alternating series test seems like a good idea. So if I prove that $\frac{1}{n+\frac{10100}{n}}$ is monotone for large $n$ and converges to $0$ then I'm done. Convergence to $0$ is trivial so all I need to do is to prove that the sequence is monotone, for sufficiently large $n$. Here is what I've done:

$\frac{1}{n+1+\frac{10100}{n+1}}-\frac{1}{n+\frac{10100}{n}}=\frac{n+\frac{10100}{n}-n-1-\frac{10100}{n+1}}{(n+1+\frac{10100}{n+1})(n+\frac{10100}{n})}$

The denominator is always positive so I'm only examining the numerator now.

$n+\frac{10100}{n}-n-1-\frac{10100}{n+1}=\frac{10100(n+1)-10100n-n(n+1)}{n(n+1)}$

Again, the denominator is always positive, so let's examine the numerator.

$10100(n+1)-10100n-n(n+1)=10100-n^2-n=10100-(n^2+n)$ which is less than zero for sufficiently large $n$. That means that for sufficiently large $n$ the sequence is monotonely decreasing so we can finish by saying our series is convergent by alternating series test.

Is everything correct with my logic? Could this be done simpler?

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    $\begingroup$ It is correct. I think it is also the simplest way to do. You can compute the derivative of $x/(x^2+10100)$ and study its sign (but is essentially the same). $\endgroup$ – Milly Dec 8 '14 at 17:39
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The logic is fine. I would look at the denominators $d_k$. If $n\gt 10100$, we have $d_n\lt n+1$. But $d_{n+1}\gt n+1$. So after $10100$ (the absolute values of) the terms are decreasing. Quicker, and a lot easier to type!

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