To which series does it relate to?

Calculus Level 4

The value of the infinite series $\sum _{ n=0 }^{ \infty } \frac{1}{{2^{n+1}}{(n+1)}} = \frac{1}{2}+\frac{1}{8}+\frac{1}{24}+\frac{1}{64}+\text{ }...$ is (give your answer with 4 significant digits):

The answer is 0.6931.

1 solution

Krishna Sharma
Oct 21, 2015

Observe that the $n^{th}$ term looks like $\dfrac{x^{n+1}}{n+1}$ So

We will start with the infinite G.P series

$\displaystyle \sum_{n=0}^{\infty} x^n = \dfrac{1}{1-x}$

And then integrating it(from 0 to x) to get

$\displaystyle \sum_{n=0}^{\infty} \dfrac{x^{n+1}}{n+1} = -ln(1-x)$

Substituting $x = \dfrac{1}{2}$ to get

$\displaystyle \boxed{\sum_{n=0}^{\infty} \dfrac{1}{2^{n+1}(n+1)} = ln2 }$

Simply do it like this... S = \sum_{n=1}^(\inf) frac{1}{(2^n) \ times (n)}

As we can see , it is written in the form of \sum {n=1}^(\inf) frac{x^n}{n} which is clearly in the form of \Li (1) (x) and \Li_(1) (x) = - \ log(1-x) Put x = frac{1}{2} , we get ,

S = - \ log(1/2)

(OR)

S = log(2)

                      (Q.E.D)

Shivam Sharma - 4 years, 2 months ago

To which series does it relate to?

The answer is 0.6931.

1 solution

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