Question

The value of $\left[\sum_{n=1}^{\infty} \frac{1}{(n+1)\sqrt{n}}\right]$ (where [.] denotes greatest integer function) is equal to

Answer 1

1

Answer 2

To find the value of $\left[\sum_{n=1}^{\infty} \frac{1}{(n+1)\sqrt{n}}\right]$, we need to evaluate the sum $S = \sum_{n=1}^{\infty} \frac{1}{(n+1)\sqrt{n}}$.

We can approximate the sum using the integral test. Since $f(x) = \frac{1}{(x+1)\sqrt{x}}$ is positive, continuous, and decreasing for $x \ge 1$, we have:

$\int_{1}^{\infty} f(x) dx \le \sum_{n=1}^{\infty} f(n) \le f(1) + \int_{1}^{\infty} f(x) dx$

Evaluating the integral:

$\int_{1}^{\infty} \frac{dx}{(x+1)\sqrt{x}}$.

Let $\sqrt{x} = t \implies x = t^2 \implies dx = 2t dt$.

When $x=1, t=1$. When $x \to \infty, t \to \infty$.

$\int_{1}^{\infty} \frac{2t dt}{(t^2+1)t} = \int_{1}^{\infty} \frac{2 dt}{t^2+1} = [2 \arctan(t)]_1^\infty = 2(\frac{\pi}{2} - \frac{\pi}{4}) = 2(\frac{\pi}{4}) = \frac{\pi}{2}$.

$f(1) = \frac{1}{(1+1)\sqrt{1}} = \frac{1}{2}$.

So, the bounds are:

$\frac{\pi}{2} \le S \le \frac{1}{2} + \frac{\pi}{2}$.

Numerically, $\pi \approx 3.14159$.

$\frac{\pi}{2} \approx 1.5708$.

$S \le 0.5 + 1.5708 = 2.0708$.

So, $1.5708 \le S \le 2.0708$.

A more accurate approximation can be obtained by computing the first few terms of the series and then approximating the tail with an integral.

$a_1 = \frac{1}{2\sqrt{1}} = \frac{1}{2} = 0.5$.

$a_2 = \frac{1}{(2+1)\sqrt{2}} = \frac{1}{3\sqrt{2}} = \frac{\sqrt{2}}{6} \approx \frac{1.414}{6} \approx 0.2357$.

$a_3 = \frac{1}{(3+1)\sqrt{3}} = \frac{1}{4\sqrt{3}} = \frac{\sqrt{3}}{12} \approx \frac{1.732}{12} \approx 0.1443$.

$a_4 = \frac{1}{(4+1)\sqrt{4}} = \frac{1}{5 \cdot 2} = \frac{1}{10} = 0.1$.

Sum of first 4 terms $\approx 0.5 + 0.2357 + 0.1443 + 0.1 = 0.98$.

$\int_{5}^{\infty} \frac{dx}{(x+1)\sqrt{x}} = 2 \left[ \arctan(\sqrt{x}) \right]_5^{\infty} = 2 \left( \frac{\pi}{2} - \arctan(\sqrt{5}) \right) \approx 2 \left( \frac{\pi}{2} - 1.150 \right) \approx 2(1.5708 - 1.150) = 2(0.4208) = 0.8416$

$S \approx 0.98 + 0.8416 = 1.8216$

Since $1.5708 \le S \le 2.0708$ and $S \approx 1.8216$, the greatest integer function $[S]$ is 1.

$[S] = 1$.