Question

Sum of first 'n' terms of a series $a_1 + a_2 + ... + a_n$ is given by $S_n = \frac{n(n^2-1)(n+2)}{4}$, then the value of $\lim_{n \to \infty} \sum_{r=2}^{n} \frac{1}{a_r}$ is

• A. 4
• B. 2
• C. $\frac{1}{4}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

$\frac{1}{4}$

Answer 2

The sum of the first 'n' terms of the series is given by $S_n = \frac{n(n^2-1)(n+2)}{4}$. We can rewrite $S_n$ as $S_n = \frac{n(n-1)(n+1)(n+2)}{4}$.

The nth term of the series, $a_n$, for $n \ge 2$ is given by $a_n = S_n - S_{n-1}$.

$a_n = \frac{n(n-1)(n+1)(n+2)}{4} - \frac{(n-1)((n-1)^2-1)((n-1)+2)}{4}$

$a_n = \frac{n(n-1)(n+1)(n+2)}{4} - \frac{(n-1)(n^2-2n)(n+1)}{4}$

$a_n = \frac{n(n-1)(n+1)(n+2)}{4} - \frac{n(n-1)(n-2)(n+1)}{4}$

Factor out the common terms $\frac{n(n-1)(n+1)}{4}$:

$a_n = \frac{n(n-1)(n+1)}{4} [(n+2) - (n-2)]$

$a_n = \frac{n(n-1)(n+1)}{4} [n+2 - n+2]$

$a_n = \frac{n(n-1)(n+1)}{4} [4]$

$a_n = n(n-1)(n+1)$ for $n \ge 2$.

The sum we need to evaluate is $\sum_{r=2}^{n} \frac{1}{a_r}$. For $r \ge 2$, $a_r = r(r-1)(r+1)$.

We need to evaluate the sum $T_n = \sum_{r=2}^{n} \frac{1}{r(r-1)(r+1)}$.

We use partial fraction decomposition for the term $\frac{1}{(r-1)r(r+1)}$.

$\frac{1}{(r-1)r} - \frac{1}{r(r+1)} = \frac{(r+1) - (r-1)}{(r-1)r(r+1)} = \frac{2}{(r-1)r(r+1)}$.

So, $\frac{1}{(r-1)r(r+1)} = \frac{1}{2} \left[ \frac{1}{(r-1)r} - \frac{1}{r(r+1)} \right]$.

Now we can write the sum $T_n$:

$T_n = \sum_{r=2}^{n} \frac{1}{2} \left[ \frac{1}{(r-1)r} - \frac{1}{r(r+1)} \right]$

$T_n = \frac{1}{2} \sum_{r=2}^{n} \left[ \frac{1}{(r-1)r} - \frac{1}{r(r+1)} \right]$

This is a telescoping sum. The sum is $\frac{1}{2} \sum_{r=2}^{n} [g(r) - g(r+1)]$.

Summing these terms, the intermediate terms cancel out:

$\sum_{r=2}^{n} [g(r) - g(r+1)] = [g(2) - g(3)] + [g(3) - g(4)] + ... + [g(n) - g(n+1)] = g(2) - g(n+1)$.

$g(2) = \frac{1}{(2-1)2} = \frac{1}{1 \cdot 2} = \frac{1}{2}$.

$g(n+1) = \frac{1}{((n+1)-1)(n+1)} = \frac{1}{n(n+1)}$.

So, the sum $T_n = \frac{1}{2} [g(2) - g(n+1)] = \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{n(n+1)} \right]$.

We need to find the limit of this sum as $n \to \infty$:

$\lim_{n \to \infty} T_n = \lim_{n \to \infty} \frac{1}{2} \left[ \frac{1}{2} - \frac{1}{n(n+1)} \right]$.

As $n \to \infty$, $n(n+1) \to \infty$, so $\frac{1}{n(n+1)} \to 0$.

$\lim_{n \to \infty} T_n = \frac{1}{2} \left[ \frac{1}{2} - 0 \right] = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.

The final answer is $\frac{1}{4}$.