1. ## series

Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman

Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))

2. Originally Posted by kingman
Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman

Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
1)

$\displaystyle \displaystyle\huge\frac{n}{(n+1)!}=\frac{n+1-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}$

$\displaystyle =\displaystyle\huge\frac{n+1}{(n+1)n!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$

Therefore

$\displaystyle \displaystyle\huge\frac{1}{2!}+\frac{2}{3!}+......$

$\displaystyle =1-\displaystyle\huge\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}+.......-\frac{1}{(n+1)!}$

$\displaystyle =1-\displaystyle\huge\frac{1}{(n+1)!}$

3. Originally Posted by kingman
Dear Sir
I would be grateful if you can help me to find f(r) in the below questions
thanks
Kingsman

Sum to the n terms of the series using difference method of telescope method .(hint Express the nth term = f(r)-f(r+1))
2)

$\displaystyle \displaystyle\huge\frac{1}{(n+2)n!}=\frac{(n+1)}{( n+2)(n+1)n!}=\frac{n+1}{(n+2)!}$

$\displaystyle \displaystyle\huge\frac{n+2-1}{(n+2)!}=\frac{n+2}{(n+2)!}-\frac{1}{(n+2)!}$

$\displaystyle \displaystyle\huge\frac{n+2}{(n+2)(n+1)!}-\frac{1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$

Use this to form the telescope.

4. Thanks very much for the kind help.
May I know how and what is the idea which you employ in your solution making it so neat and simply. Is the f(r) unique and is there a general method to find f(r) ?.
Lastly is it possible to find a g(r) such that the nth term= g(r)-g(r-1) allowing us to find the sum?

5. If you are told to evaluate an infinite sum you should either change it into a known form of a function or you should make it a telescoping sum. Now to make it a telescoping sum you need to look for the following things:

- an obvious way to apply partial fractions decomposition
- add and subtract something in the numerator so that part of it looks like something that we can cancel on the bottom.

That is usually a good start. There is a lot of intuition involved in some examples and less in others.

6. Thanks very much for the reply.
Can I know how to find g(r) such that the nth term= g(r)-g(r-1) thus allowing us to find the sum of the above 3 problem.
How to apply telescope method in question 3?
I cannot wrap around my head to get started in problem3
Thanks

7. 3)

The sum can be written as...

$\displaystyle \displaystyle \sum_{k=1}^{n} \frac{k}{k+5} = n - 5\ \sum_{k=1}^{n} \frac{1}{k+5}$ (1)

In order to solve the second term of (1) we have to introduce the function digamma, defined as...

$\displaystyle \displaystyle \psi(x)= \sum_{j=1}^{\infty} \frac{x}{j\ (j+x)} - \gamma$ (2)

... where $\displaystyle \gamma$ is the 'Euler's constant'. From (2) You can derive the following basic identity...

$\displaystyle \displaystyle \psi(1+x) - \psi (x) = \frac{1}{1+x}$ (3)

Now setting in (3) $\displaystyle x=k+4$ You obtain...

$\displaystyle \displaystyle \psi(k+5) - \psi (k+4) = \frac{1}{k+5}$ (4)

... so that is...

$\displaystyle \displaystyle \sum_{k=1}^{n} \frac{1}{k+5} = \sum_{k=1}^{n} \{ \psi(k+5) - \psi (k+4) \} = \psi (n+5) - \psi(5)$ (5)

... and finally You can write...

$\displaystyle \displaystyle \sum_{k=1}^{n} \frac{k}{k+5} = n - 5\ \psi(n+5) + 5\ \psi(5)$ (6)

Digamma function - Wikipedia, the free encyclopedia

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

8. That's pretty cool, chisigma, but I think there is an even simpler answer.

$\displaystyle \displaystyle\sum_{k=1}^n \frac{k}{k+5} = \sum_{k=1}^n \frac{k+5-5}{k+5} = \sum_{k=1}^n \left(\frac{k+5}{k+5}-\frac{5}{k+5}\right)$

$\displaystyle \displaystyle = \left(\sum_{k=1}^n 1\right)-5\left(\sum_{k=1}^n \frac{1}{k+5}\right) = n - 5\sum_{k=6}^{n+5} \frac{1}{k+5} = n - 5\left(H_n - \frac{1}{1}-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right) = \frac{137}{12}+n-5H_n$

where $\displaystyle H_n$ is the n-th Harmonic number. There is no simpler way to write this sum.