Is there exists a function like this

Calculus Level 2

If $\displaystyle f(x) = \sum_{k=1}^\infty \frac x{\left((k-1)x+1\right)(kx+1)}$ , what is the value of $f(2017) + f\left(\dfrac 1{2017}\right)$ ?

The answer is 2.

1 solution

Chew-Seong Cheong
Mar 1, 2018

$\begin{aligned} f(x) & = \sum_{\color{#3D99F6}k=1}^\infty \frac x{((k-1)x+1)(kx+1)} \\ & = \sum_{\color{#D61F06}k=0}^\infty \frac x{(kx+1)((k+1)x+1)} \\ & = \sum_{k=0}^\infty \left(\frac 1{kx+1} - \frac 1{kx+x+1} \right) \\ & = 1 \end{aligned}$

Therefore, $f(2017) + f\left(\dfrac 1{2017}\right) = 1 + 1 = \boxed{2}$ .

@Ayush Mishra , it is better not to use $i$ which can be mistaken as $i=\sqrt {-1}$ , the imaginary unit, and $\{ \cdot \}$ , which can be mistaken as fractional part function. I have edited the problem for you, since I am the first solver.

Since you like to post problems, it is advisable to learn up LaTex to enter the codes directly instead of using the LaTex editor here. You don't need to enter so many { } and \left( \right). You can see the LaTex codes by placing the mouse cursor on the formulas. Our click the pull-down menu " $\cdots$ More" at the bottom of the answer section and select "Toggle LaTex".

Chew-Seong Cheong - 3 years, 3 months ago

Thank you sir

A Former Brilliant Member - 3 years, 3 months ago

Telescoping helps !

Vijay Simha - 1 year, 7 months ago

Is there exists a function like this

The answer is 2.

1 solution

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