To infinity and beyond

Calculus Level 3

$\dfrac{1}{1 + 1^2 + 1^4} + \dfrac{2}{ 1 + 2^2 + 2^4} + \dfrac{3}{1 + 3^2 + 3^4} + \ldots + \dfrac{n}{1 + n^2 + n^4}$

Let $S_n$ denote summation above. Evaluate $\displaystyle \lim_{n \to \infty} S_n$ .

The answer is 0.5.

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Brian Charlesworth
Nov 28, 2014

Note first that, for any positive integer $k$ ,

$\dfrac{k}{1 + k^{2} + k^{4}} = \dfrac{k}{(k^{2} + 1)^{2} - k^{2}} = \dfrac{k}{(k^{2} - k + 1)(k^{2} + k + 1)} =$

$(\dfrac{1}{2})*(\dfrac{1}{k^{2} - k + 1} - \dfrac{1}{k^{2} + k + 1})$ .

Next, note that

$\dfrac{1}{(k + 1)^{2} - (k + 1) + 1} = \dfrac{1}{k^2 + k + 1}$ .

As a result, $S_{n} = \displaystyle\sum_{k=1}^{n} \dfrac{k}{1 + k^{2} + k^{4}} = (\dfrac{1}{2})*\displaystyle\sum_{k=0}^{n} (\dfrac{1}{k^{2} - k + 1} - \dfrac{1}{k^{2} + k + 1})$

is a telescoping sum, (i.e., all the "interior" terms cancel each other out). This leaves us with

$S_{n} = (\dfrac{1}{2})*(\dfrac{1}{1 - 1^{2} + 1^{4}} - \dfrac{n}{1 + n^{2} + n^{4}}) = \dfrac{1}{2} - (\dfrac{1}{2})*\dfrac{n}{1 + n^{2} + n^{4}}$ ,

and so $\lim_{n \rightarrow \infty} S_{n} = \dfrac{1}{2} = \boxed{0.5}$ .

Can u use L-hopital rule to solve?

Loki Come - 6 years, 6 months ago

L'Hopital's doesn't really apply here as we are working with a discrete variable and $S_{n}$ is not in an indeterminate form as $n \longrightarrow \infty$ .

Brian Charlesworth - 6 years, 6 months ago

But why L'Hopital's doesnt work in discrete variable?

Loki Come - 6 years, 6 months ago

@Loki Come – L'Hopital's rule is applicable only to differentiable functions, and differentiability implies that the function is continuous. By definition, discrete functions are not continuous, and hence, strictly speaking, L'Hopital's rule cannot be applied directly to a discrete function.

However, if the discrete function can be "mapped" to a continuous and differentiable function, then we can apply L'Hopital's rule to this analogous function and attribute the subsequent results to the discrete function as well.

There is also a sort of "L'Hopital's rule for discrete variables" known as the Stolz-Cesaro Theorem , which might come in useful. :)

Brian Charlesworth - 6 years, 6 months ago

@Brian Charlesworth – This took me way too long... I was trying to heaviside the two quadratics. Lol

Trevor Arashiro - 6 years, 6 months ago

@Trevor Arashiro – Well, I'm impressed that you know about the Heaviside method, in any event. :)

Brian Charlesworth - 6 years, 6 months ago

@Brian Charlesworth – Is it that high level? Because my friend taught it to me last year (he's way smarter than I).

Trevor Arashiro - 6 years, 6 months ago

@Trevor Arashiro – Maybe not, but in my experience partial fraction decomposition is dealt with in an introductory calculus class, which would be either in grade 12 or first-year university.

Brian Charlesworth - 6 years, 6 months ago

To infinity and beyond

The answer is 0.5.

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