Question

$\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\frac{3}{1+3^2+3^4}+...=H/w$

Answer 1

1/2

Answer 2

The problem asks to find the sum of the infinite series:

$S = \sum_{n=1}^{\infty} \frac{n}{1+n^2+n^4}$

Let's analyze the general term of the series, $T_n = \frac{n}{1+n^2+n^4}$.

Step 1: Factorize the denominator

The denominator $1+n^2+n^4$ can be factorized using the identity $a^4+a^2+1 = (a^2+1)^2 - a^2 = (a^2+1-a)(a^2+1+a)$. So, $1+n^2+n^4 = (n^2-n+1)(n^2+n+1)$.

Step 2: Rewrite the general term as a difference of two terms

Substitute the factorization back into $T_n$:

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

Notice that the difference between the two factors in the denominator is $(n^2+n+1) - (n^2-n+1) = 2n$. To utilize this, multiply and divide $T_n$ by 2:

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

Now, replace $2n$ with the difference of the factors:

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

This can be split into two fractions:

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

Step 3: Identify the telescoping sum pattern

Let $f(n) = \frac{1}{n^2-n+1}$. Then, let's evaluate $f(n+1)$:

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

So, the general term $T_n$ can be written as:

$T_n = \frac{1}{2} (f(n) - f(n+1))$

Step 4: Calculate the partial sum $S_N$

The sum of the first $N$ terms is $S_N = \sum_{n=1}^{N} T_n$:

$S_N = \sum_{n=1}^{N} \frac{1}{2} (f(n) - f(n+1))$ $S_N = \frac{1}{2} [(f(1) - f(2)) + (f(2) - f(3)) + (f(3) - f(4)) + \dots + (f(N) - f(N+1))]$

This is a telescoping sum where intermediate terms cancel out:

$S_N = \frac{1}{2} [f(1) - f(N+1)]$

Step 5: Evaluate $f(1)$ and $f(N+1)$

$f(1) = \frac{1}{1^2-1+1} = \frac{1}{1} = 1$ $f(N+1) = \frac{1}{(N+1)^2-(N+1)+1} = \frac{1}{N^2+N+1}$

Substitute these values back into $S_N$:

$S_N = \frac{1}{2} \left[ 1 - \frac{1}{N^2+N+1} \right]$

Step 6: Find the sum of the infinite series

The sum of the infinite series $S$ is the limit of $S_N$ as $N \to \infty$:

$S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \frac{1}{2} \left[ 1 - \frac{1}{N^2+N+1} \right]$

As $N \to \infty$, the term $\frac{1}{N^2+N+1}$ approaches 0.

$S = \frac{1}{2} [1 - 0] = \frac{1}{2}$

The sum of the given series is $\frac{1}{2}$.