# How do you find the limit of mathematical series ?

Nov 11, 2016

That depends on the series. There are several different kinds of techniques we can use for the different types of series.

#### Explanation:

Do you have any examples?

Nov 12, 2016

This series does not converge (it does not have a limit).

#### Explanation:

The series in question is the sum of $\frac{1}{\sqrt{2 + n}}$ from $n = 1$ to $\infty$, which can be expressed in summation notation as:
${\sum}_{n = 1}^{\infty} \frac{1}{\sqrt{2 + n}}$

The first few terms are:
$\frac{1}{\sqrt{2 + 1}} + \frac{1}{\sqrt{2 + 2}} + \frac{1}{\sqrt{2 + 3}} + \frac{1}{\sqrt{2 + 4}} \ldots$
$= \frac{1}{\sqrt{3}} + \frac{1}{2} + \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{6.}} . .$

At first glance, the series seems to converge - which means after you take all of its terms and add them up, you get a number. However, take a look at this series:
${\sum}_{n = 1}^{\infty} \frac{1}{n}$

The first few terms are:
$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4.} . .$

This is called the harmonic series, and it is known to diverge - i.e. the sum keeps getting bigger and bigger as $n$ goes to $\infty$.

Compare the first few terms of the harmonic series with our series:
$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4.} . .$
$\frac{1}{\sqrt{3}} + \frac{1}{2} + \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{6.}} . .$

Our series is bigger than the harmonic series after $\frac{1}{2}$ ($\frac{1}{\sqrt{5}} > \frac{1}{3}$, $\frac{1}{\sqrt{6}} > \frac{1}{4}$, in general $\frac{1}{\sqrt{n + 2}} > \frac{1}{n}$ for $n > 2$).

What this means is that our series is larger than the divergent harmonic series, so the sum of our series should be larger than the sum of the harmonic series . But that means our series diverges too - it will just get bigger and bigger as $n$ goes to $\infty$. Thus, the limit of this series is $\infty$ - in other words, it does not exist. Keep in mind that this isn't an especially rigorous proof, but it works.