Can This Infinite Series Be Summed?

[anemone]

Evaluate $\dfrac{1}{3^2+1}+\dfrac{1}{4^2+2}+\dfrac{1}{5^2+3}+\cdots$

 

[MarkFL]

My solution:

Spoiler

\(\displaystyle S=\sum_{k=1}^{\infty}\left(\frac{1}{(k+2)^2+k} \right)=\sum_{k=1}^{\infty}\left(\frac{1}{(k+1)(k+4)} \right)\)

Using partial fraction decomposition on the summand, we find:

\(\displaystyle S=\frac{1}{3}\sum_{k=1}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k+4} \right)\)

We may write this as:

\(\displaystyle S=\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\sum_{k=1}\left(\frac{1}{k+4}-\frac{1}{k+4} \right) \right)\)

And so we have:

\(\displaystyle S=\frac{1}{3}\cdot\frac{13}{12}=\frac{13}{36}\)

 

[anemone]

My solution:

Spoiler

\(\displaystyle S=\sum_{k=1}^{\infty}\left(\frac{1}{(k+2)^2+k} \right)=\sum_{k=1}^{\infty}\left(\frac{1}{(k+1)(k+4)} \right)\)

Using partial fraction decomposition on the summand, we find:

\(\displaystyle S=\frac{1}{3}\sum_{k=1}^{\infty}\left(\frac{1}{k+1}-\frac{1}{k+4} \right)\)

We may write this as:

\(\displaystyle S=\frac{1}{3}\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\sum_{k=1}\left(\frac{1}{k+4}-\frac{1}{k+4} \right) \right)\)

And so we have:

\(\displaystyle S=\frac{1}{3}\cdot\frac{13}{12}=\frac{13}{36}\)

Hey MarkFL, your method is so elegant and neat! Well done, my sweet admin!

 

[Saitama]

Evaluate $\dfrac{1}{3^2+1}+\dfrac{1}{4^2+2}+\dfrac{1}{5^2+3}+\cdots$

Spoiler

Notice that the given sum can be written as:
$$\sum_{r=1}^{\infty} \frac{1}{(r+2)^2+r}=\sum_{r=1}^{\infty} \frac{1}{r^2+5r+4}=\sum_{r=1}^{\infty} \frac{1}{(r+4)(r+1)}$$
$$=\frac{1}{3}\left(\sum_{r=1}^{\infty} \frac{1}{r+1}-\frac{1}{r+4}\right)$$
$$=\frac{1}{3}\left(\sum_{r=1}^{\infty}\int_0^1 x^r-x^{r+3}\,dx\right)=\frac{1}{3}\left( \sum_{r=1}^{\infty} \int_0^1 x^r(1-x^3)\,dx\right)$$
$$=\frac{1}{3}\int_0^1 (1-x^3)\frac{x}{1-x}\,dx = \frac{1}{3}\int_0^1 x(x^2+x+1)\,dx=\frac{1}{3}\int_0^1 x^3+x^2+x \,dx$$
Evaluating the definite integral gives:
$$\frac{1}{3}\cdot \frac{13}{12}=\frac{13}{36}$$

 

[anemone]

Spoiler

Notice that the given sum can be written as:
$$\sum_{r=1}^{\infty} \frac{1}{(r+2)^2+r}=\sum_{r=1}^{\infty} \frac{1}{r^2+5r+4}=\sum_{r=1}^{\infty} \frac{1}{(r+4)(r+1)}$$
$$=\frac{1}{3}\left(\sum_{r=1}^{\infty} \frac{1}{r+1}-\frac{1}{r+4}\right)$$
$$=\frac{1}{3}\left(\sum_{r=1}^{\infty}\int_0^1 x^r-x^{r+3}\,dx\right)=\frac{1}{3}\left( \sum_{r=1}^{\infty} \int_0^1 x^r(1-x^3)\,dx\right)$$
$$=\frac{1}{3}\int_0^1 (1-x^3)\frac{x}{1-x}\,dx = \frac{1}{3}\int_0^1 x(x^2+x+1)\,dx=\frac{1}{3}\int_0^1 x^3+x^2+x \,dx$$
Evaluating the definite integral gives:
$$\frac{1}{3}\cdot \frac{13}{12}=\frac{13}{36}$$

Hmm...another good method to solve this problem, thanks Pranav for the solution and also for participating!:)

 

[Saitama]

Hmm...another good method to solve this problem, thanks Pranav for the solution and also for participating!:)

You should thank Random Variable for this. :p

http://mathhelpboards.com/calculus-10/limit-sum-8576.html#post39742

 

[anemone]

You should thank Random Variable for this. :p

http://mathhelpboards.com/calculus-10/limit-sum-8576.html#post39742

I will...but, does this mean you are kind of "cheating" here? Hehehe...just kidding!:p