Question

I'm trying some summation this afternoon, I was trying to add up positive odd numbers.

So 1+3+5+7+9+11+13... etc. etc.
Odd number could be written algebraically as 2x+1, x being any positive natural number.

\[\huge \sum_{x=0}^{n} (2x+1)\]

And this is what I got:
\[\huge \sum_{x=0}^{n} (2x+1)=(n-1)^2, n \in \Natural\]

Answer 1

Now I'm trying \[\huge \sum_{x=0}^{n}\frac{1}{2x+1}\]

Answer 2

but I can't find how to write that in with n as variable :|

Answer 3

So in short, you're trying to find a simpler expression for \[\sum_{x=0}^n \frac{1}{2x+1}\]
in terms of n but you can't find one, right?

Answer 4

Yup :)

Answer 5

Not possible without using some advance function.

Answer 6

And your earlier answer is incorrect,

\[ \sum_{x=0}^{n} (2x+1)=(n\color{red}{+}1)^2, n \in \mathbb{N} \]

Answer 7

Oh, I should check my work again D:

Answer 8

To be more pedantic:

\[ \sum_{x=0}^{n} (2x+1)=(n\color{red}{+}1)^2, n \in \mathbb{N_0} \]

Answer 9

Let \(H_n\) be the nth harmonic number. http://en.wikipedia.org/wiki/Harmonic_number

\[\begin{align}1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1}&=H_{2n+2}-\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2n}\right)\\
&=H_{2n+2}-\frac{H_n}{2}
\end{align}\]

This was the first thing I thought of.