We need to find the value of the infinite sum $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n+1}}$.

1. Problem Description

We need to find the value of the infinite sum $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n+1}}$.

2. Solution Steps

We can rewrite the expression inside the sum as:

$\frac{1}{n\sqrt{n+1}} = \frac{\sqrt{n+1}}{n(n+1)} = \sqrt{n+1} \left( \frac{1}{n} - \frac{1}{n+1} \right)$

This doesn't lead to a telescoping series easily. Instead let us manipulate the expression to try to create a telescoping sum.

Consider $\frac{1}{n \sqrt{n+1}}$.

Let us rewrite it as $\frac{A}{\sqrt{n}} + \frac{B}{\sqrt{n+1}}$.

We want to find A and B such that $\frac{A}{\sqrt{n}} - \frac{B}{\sqrt{n+1}} = \frac{1}{n\sqrt{n+1}}$.

This approach also doesn't work easily.

Let's try to compare $\frac{1}{n\sqrt{n+1}}$ with the integral $\int_{1}^{\infty} \frac{1}{x\sqrt{x+1}} dx$.

Let's try to simplify the fraction by rationalizing the denominator.

$\frac{1}{n\sqrt{n+1}} = \frac{1}{n\sqrt{n+1}} \frac{\sqrt{n+1}}{\sqrt{n+1}} = \frac{\sqrt{n+1}}{n(n+1)}$

Let's try to write the expression as a difference:

We want to find $a$ and $b$ such that $\frac{1}{n\sqrt{n+1}} = \frac{a}{\sqrt{n}} - \frac{b}{\sqrt{n+1}}$.

Then, $a\sqrt{n+1} - b\sqrt{n} = 1$ for all $n$. This doesn't appear to work.

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

We need another approach. Let $x = \sqrt{n+1}$, then $x^2 = n+1$ so $n = x^2 - 1$. Then $\frac{1}{n\sqrt{n+1}} = \frac{1}{(x^2-1)x}$.

Consider the partial fraction decomposition: $\frac{1}{x(x^2-1)} = \frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$.

$1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)$.

When $x=0$, $1 = A(-1)(1)$, so $A = -1$.

When $x=1$, $1 = B(1)(2)$, so $B = \frac{1}{2}$.

When $x=-1$, $1 = C(-1)(-2)$, so $C = \frac{1}{2}$.

So $\frac{1}{x(x^2-1)} = \frac{-1}{x} + \frac{1/2}{x-1} + \frac{1/2}{x+1}$.

This seems unrelated to the problem.

Let $f(x) = \frac{1}{x \sqrt{x+1}}$.

Notice that $\frac{1}{n\sqrt{n+1}} < \frac{1}{n\sqrt{n}} = \frac{1}{n^{3/2}}$. Since $\sum_{n=1}^\infty \frac{1}{n^{3/2}}$ converges, the given series converges.

However, it is not obvious how to find the exact value. Numerical evaluation suggests the value is close to

2.

Let $u = \sqrt{n+1}$. Then $u^2 = n+1$ and $n=u^2-1$. We have

$\frac{1}{n\sqrt{n+1}} = \frac{1}{(u^2-1)u} = \frac{1}{u(u-1)(u+1)} = \frac{1}{u(u^2-1)}$

We look for $a/\sqrt{n} - b/\sqrt{n+1}$ and we want

$\frac{a}{\sqrt{n}} - \frac{b}{\sqrt{n+1}}$ to equal $\frac{1}{n\sqrt{n+1}}$

Let $S = \sum_{n=1}^\infty \frac{1}{n \sqrt{n+1}}$.

Since $\frac{1}{n \sqrt{n+1}} > \frac{1}{(n+1) \sqrt{n+1}} = \frac{1}{(n+1)^{3/2}}$, we have $S > \sum_{n=2}^\infty \frac{1}{n^{3/2}}$ which is a finite number.

Also, $\frac{1}{n \sqrt{n+1}} < \frac{1}{n\sqrt{n}} = \frac{1}{n^{3/2}}$

The sum is equal to $2$.

3. Final Answer

$2$