Question

The sum to infinity of the series $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}$ +...... is equal to -

• A. 2
• B. 5/2
• C. 3
• D. none of these

Answer 1

Answer: (A)

2

Answer 2

The given series is $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}$ +......

First, let's find the general term of the series, $T_n$. The denominator of the $n$-th term is the sum of the first $n$ natural numbers, which is given by the formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.

So, the $n$-th term of the series is: $T_n = \frac{1}{\frac{n(n+1)}{2}} = \frac{2}{n(n+1)}$

To find the sum to infinity, we need to evaluate $S_\infty = \sum_{n=1}^{\infty} T_n$. We can express $T_n$ using partial fraction decomposition: $T_n = \frac{2}{n(n+1)} = 2 \left( \frac{1}{n} - \frac{1}{n+1} \right)$

Now, let's write out the partial sum $S_N = \sum_{n=1}^{N} T_n$: $S_N = \sum_{n=1}^{N} 2 \left( \frac{1}{n} - \frac{1}{n+1} \right)$ $S_N = 2 \left[ \left( \frac{1}{1} - \frac{1}{1+1} \right) + \left( \frac{1}{2} - \frac{1}{2+1} \right) + \left( \frac{1}{3} - \frac{1}{3+1} \right) + \dots + \left( \frac{1}{N} - \frac{1}{N+1} \right) \right]$ $S_N = 2 \left[ \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \dots + \left( \frac{1}{N} - \frac{1}{N+1} \right) \right]$

This is a telescoping series, where intermediate terms cancel out. $S_N = 2 \left[ 1 - \frac{1}{N+1} \right]$

Finally, to find the sum to infinity, we take the limit as $N \to \infty$: $S_\infty = \lim_{N \to \infty} S_N = \lim_{N \to \infty} 2 \left[ 1 - \frac{1}{N+1} \right]$ As $N \to \infty$, $\frac{1}{N+1}$ approaches 0. $S_\infty = 2 \left[ 1 - 0 \right] = 2 \times 1 = 2$

The sum to infinity of the series is 2.