Limits vs Summation

Calculus Level 3

$\large \displaystyle\lim_{n\to\infty}\left(\dfrac{n^2b+1}{2n^2a-1}\right)=\ 1$

Suppose for real numbers $a,b$ , we have the limit above.

What is the value of $\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{ab}{a^2+b^2}\right)^n \ ?$

The answer is 0.667.

1 solution

Вук Радовић
May 9, 2015

Using L'Hospital's Rule you will easy find that

$b=2a$ and

$\frac{ab}{a^2+b^2} = \frac{2}{5}$

So the sum is equal $\frac{2}{3}$

Moderator note:

You don't need to implore that rule. Hint: If the limit is finite, what can we say about the coefficients of the leading terms of the numerator and denominator?

hey is it pronounced as L'Hospital in your country as well........I thought that is the case in India only. ( Actually it's L'opital's rule.)

Kislay Raj - 6 years, 1 month ago

Oh sorry for mistake. Actually i heard for this rule on this site im 17 and in my country we are learning logorithm's right now so i haven't got so much chance to hear things about calculus. :/

Вук Радовић - 6 years, 1 month ago

Limits vs Summation

The answer is 0.667.

1 solution

Moderator note:

0 pending reports