MHB Can This Infinite Series Be Summed?

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Infinity Sum
AI Thread Summary
The discussion centers on evaluating the infinite series $\dfrac{1}{3^2+1}+\dfrac{1}{4^2+2}+\dfrac{1}{5^2+3}+\cdots$. Participants express appreciation for elegant solutions, particularly highlighting contributions from members like MarkFL and Pranav. There is a light-hearted exchange regarding the use of external resources, with a mention of Random Variable's influence on the problem-solving process. The conversation maintains a friendly tone while focusing on the mathematical evaluation. Overall, the thread emphasizes collaboration and sharing of methods in tackling the series.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Evaluate $\dfrac{1}{3^2+1}+\dfrac{1}{4^2+2}+\dfrac{1}{5^2+3}+\cdots$
 
Mathematics news on Phys.org
My solution:

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

Using partial fraction decomposition on the summand, we find:

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

We may write this as:

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

And so we have:

$$S=\frac{1}{3}\cdot\frac{13}{12}=\frac{13}{36}$$
 
MarkFL said:
My solution:

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

Using partial fraction decomposition on the summand, we find:

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

We may write this as:

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

And so we have:

$$S=\frac{1}{3}\cdot\frac{13}{12}=\frac{13}{36}$$

Hey MarkFL, your method is so elegant and neat!:cool: Well done, my sweet admin!
 
anemone said:
Evaluate $\dfrac{1}{3^2+1}+\dfrac{1}{4^2+2}+\dfrac{1}{5^2+3}+\cdots$

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:
$$\frac{1}{3}\cdot \frac{13}{12}=\frac{13}{36}$$
 
Pranav said:
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!:)
 
anemone said:
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
 
Pranav said:
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
 
Back
Top