# How do you use the limit comparison test for sum 1 / (n + sqrt(n)) for n=1 to n=oo?

Aug 10, 2015

${\sum}_{n = 1}^{\infty} \frac{1}{n + \sqrt{n}}$ diverges, this can be seen by comparing it to ${\sum}_{n = 1}^{\infty} \frac{1}{2 n}$.

#### Explanation:

Since this series is a sum of positive numbers, we need to find either a convergent series ${\sum}_{n = 1}^{\infty} {a}_{n}$ such that ${a}_{n} \ge \frac{1}{n + \sqrt{n}}$ and conclude that our series is convergent, or we need to find a divergent series such that ${a}_{n} \le \frac{1}{n + \sqrt{n}}$ and conclude our series to be divergent as well.

We remark the following:
For
$n \ge 1$, $\sqrt{n} \le n$.
Therefore
$n + \sqrt{n} \le 2 n$.
So
$\frac{1}{n + \sqrt{n}} \ge \frac{1}{2 n}$.

Since it is well known that ${\sum}_{n = 1}^{\infty} \frac{1}{n}$ diverges, so ${\sum}_{n = 1}^{\infty} \frac{1}{2 n}$ diverges as well, since if it would converge, then $2 {\sum}_{n = 1}^{\infty} \frac{1}{2 n} = {\sum}_{n = 1}^{\infty} \frac{1}{n}$ would converge as well, and this is not the case.

Now using the comparison test, we see that ${\sum}_{n = 1}^{\infty} \frac{1}{n + \sqrt{n}}$ diverges.

Nov 20, 2016

The limit comparison test takes two series, $\sum {a}_{n}$ and $\sum {b}_{n}$ where ${a}_{n} \ge 0$, ${b}_{n} > 0$.

If ${\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n} = L$ where $L > 0$ and is finite, then either both series converge or both series diverge.

We should let ${a}_{n} = \frac{1}{n + \sqrt{n}}$, the sequence from the given series. A good ${b}_{n}$ choice is the overpowering function that ${a}_{n}$ approaches as $n$ becomes large. So, let ${b}_{n} = \frac{1}{n}$.

Note that $\sum {b}_{n}$ diverges (it's the harmonic series).

So, we see that ${\lim}_{n \rightarrow \infty} {a}_{n} / {b}_{n} = {\lim}_{n \rightarrow \infty} \frac{\frac{1}{n + \sqrt{n}}}{\frac{1}{n}} = {\lim}_{n \rightarrow \infty} \frac{n}{n + \sqrt{n}}$. Continuing by dividing through by $\frac{n}{n}$, this becomes ${\lim}_{n \rightarrow \infty} \frac{1}{1 + \frac{1}{\sqrt{n}}} = \frac{1}{1} = 1$.

Since the limit is $1$, which is $> 0$ and defined, we see that $\sum {a}_{n}$ and $\sum {b}_{n}$ will both diverge or converge. Since we already know at $\sum {b}_{n}$ diverges, we can conclude that $\sum {a}_{n} = {\sum}_{n = 1}^{\infty} \frac{1}{n + \sqrt{n}}$ diverges as well.