Question

How do you find the limit of mathematical series ?

Answer 1

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?

Answer 2

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

Explanation:

The series in question is the sum of `1/sqrt(2+n)` from `n=1` to `oo`, which can be expressed in summation notation as:
`sum_(n=1)^oo 1/sqrt(2+n)`

The first few terms are:
`1/sqrt(2+1)+1/sqrt(2+2)+1/sqrt(2+3)+1/sqrt(2+4)...`
`=1/sqrt3+1/2+1/sqrt5+1/sqrt6...`

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)^oo 1/n`

The first few terms are:
`1/1+1/2+1/3+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 `oo`.

Compare the first few terms of the harmonic series with our series:
`1/1+1/2+1/3+1/4...`
`1/sqrt3+1/2+1/sqrt5+1/sqrt6...`

Our series is bigger than the harmonic series after `1/2` (`1/sqrt5>1/3`, `1/sqrt6>1/4`, in general `1/sqrt(n+2)>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 `oo`. Thus, the limit of this series is `oo` - in other words, it does not exist. Keep in mind that this isn't an especially rigorous proof, but it works.