# Find the sum upto infinite terms of the series: 1/(1*3) + 2/(1*3*5) + 3/(1*3*5*7) + 4/(1*3*5*7*9)....... Using partial fractions?

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Steve M Share
Sep 6, 2017

$\frac{1}{2}$

#### Explanation:

We seek:

$S = \frac{1}{1.3} + \frac{2}{1.3 .5} + \frac{3}{1.3 .5 .7} + \frac{4}{1.3 .5 .7 .9} + \ldots$

The general term is given by:

 u_n = n/(1.3.5...(2n-1)(2n+1)

$\setminus \setminus \setminus \setminus = \frac{\frac{2 n + 1 - 1}{2}}{1.3 .5 \ldots \left(2 n - 1\right) \left(2 n + 1\right)}$

$\setminus \setminus \setminus \setminus = \frac{1}{2} \left\{\frac{2 n + 1 - 1}{1.3 .5 \ldots \left(2 n - 1\right) \left(2 n + 1\right)}\right\}$

$\setminus \setminus \setminus \setminus = \frac{1}{2} \left\{\frac{2 n + 1}{1.3 .5 \ldots \left(2 n - 1\right) \left(2 n + 1\right)} - \frac{1}{1.3 .5 \ldots \left(2 n - 1\right) \left(2 n + 1\right)}\right\}$

$\setminus \setminus \setminus \setminus = \frac{1}{2} \left\{\frac{1}{1.3 .5 \ldots \left(2 n - 1\right)} - \frac{1}{1.3 .5 \ldots \left(2 n + 1\right)}\right\}$

If we write out the first few terms using this, we get:

$n = 1 \implies {u}_{1} = \frac{1}{2} \left(\frac{1}{1} - \frac{1}{1.3}\right) \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{3}$
$n = 2 \implies {u}_{2} = \frac{1}{2} \left(\frac{1}{1.3} - \frac{1}{1.3 .5}\right) \setminus \setminus \setminus \setminus \setminus = \frac{2}{15}$
$n = 3 \implies {u}_{3} = \frac{1}{2} \left(\frac{1}{1.3 .5} - \frac{1}{1.3 .5 .7}\right) = \frac{1}{35}$

So we can write the sum of the first n terms as follows:

${S}_{n} = {u}_{1} + {u}_{2} + \ldots {u}_{n}$

$\setminus \setminus \setminus \setminus \setminus = \left\{\frac{1}{2} \left(\frac{1}{1} - \frac{1}{1.3}\right)\right\} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left\{\frac{1}{2} \left(\frac{1}{1.3} - \frac{1}{1.3 .5}\right)\right\} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left\{\frac{1}{2} \left(\frac{1}{1.3 .5} - \frac{1}{1.3 .5 .7}\right)\right\} +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \vdots$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left\{\frac{1}{2} \left(\frac{1}{1.3 .5 \ldots \left(2 n - 1\right)} - \frac{1}{1.3 .5 \ldots \left(2 n + 1\right)}\right)\right\}$

 \ \ \ \ \ = 1/2{(1/1 - 1/(1.3) ) +
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{1.3} - \frac{1}{1.3 .5}\right) +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left(\frac{1}{1.3 .5} - \frac{1}{1.3 .5 .7}\right) +$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \vdots$
 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( 1/(1.3.5...(2n-1)) - 1/(1.3.5...(2n+1)) ) }

This is a difference sum, and we can see that almost all terms will cancel with other terms, after which we are left with:

${S}_{n} = \frac{1}{2} \left\{1 - \frac{1}{1.3 .5 \ldots \left(2 n + 1\right)}\right\}$

And for the sum we seek we have:

$S = {\lim}_{n \rightarrow \infty} {S}_{n}$
$\setminus \setminus = {\lim}_{n \rightarrow \infty} \frac{1}{2} \left\{1 - \frac{1}{1.3 .5 \ldots \left(2 n + 1\right)}\right\}$
$\setminus \setminus = \frac{1}{2} \left\{1 - 0\right\}$
$\setminus \setminus = \frac{1}{2}$

Although a surprising result we can readily confirm this via an excel spreadsheet:

Then teach the underlying concepts
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#### Explanation

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3
Dec 10, 2017

$\frac{1}{2.}$

#### Explanation:

Suppose that,

${S}_{n} = \frac{1}{1 \cdot 3} + \frac{2}{1 \cdot 3 \cdot 5} + \frac{3}{1 \cdot 3 \cdot 5 \cdot 7} + \ldots + {t}_{n} .$

It is easy to see that, the Nr. of the General ${n}^{t h}$ term, i.e.,

${t}_{n} ,$ is $n ,$ whereas, its Dr. is, the Product of the first

$\left(n + 1\right)$ terms of the A.P. : $1 , 3 , 5 , 7 , \ldots \ldots ,$which is, given by,

$1 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot \left(2 n + 1\right)$.

$\therefore {t}_{n} = \frac{n}{1 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot \left(2 n + 1\right)} , n \in \mathbb{N} .$

We note that, ${t}_{n} = \frac{1}{2} \left[\frac{\left(2 n + 1\right) - 1}{1 \cdot 3 \cdot 5 \cdot 7 \cdot \ldots \cdot \left(2 n - 1\right) \left(2 n + 1\right)}\right]$,

$= \frac{1}{2} \left\{\frac{2 n + 1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot \left(2 n - 1\right) \left(2 n + 1\right)} - \frac{1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot \left(2 n + 1\right)}\right\}$

$\therefore {t}_{n} = \frac{1}{2} \left\{\frac{1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot \left(2 n - 1\right)} - \frac{1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot \left(2 n + 1\right)}\right\} .$

$\therefore {S}_{n} = {t}_{1} + {t}_{2} + {t}_{3} + \ldots + {t}_{n} ,$

=1/2[{1/1cancel(-1/(1*3))}+{cancel(1/(1*3))cancel(-1/(1*3*5))}
$+ \left\{\cancel{\frac{1}{1 \cdot 3 \cdot 5}} \cancel{- \frac{1}{1 \cdot 3 \cdot 5 \cdot 7}}\right\}$
+...+{cancel(1/(1*3*5*...*(2n-1)))-1/(1*3*5*...*(2n+1))}].

$\Rightarrow {S}_{n} = \frac{1}{2} \left[1 - \frac{1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot \left(2 n + 1\right)}\right] .$

Now, by Defn., the sum $S$ of the given infinite series, is,

$S = {\lim}_{n \to \infty} {S}_{n} ,$

$= {\lim}_{n \to \infty} \frac{1}{2} \left[1 - \frac{1}{1 \cdot 3 \cdot 5 \cdot \ldots \cdot \left(2 n + 1\right)}\right] ,$

$= \frac{1}{2} \left[1 - 0\right] ,$

$\Rightarrow S = \frac{1}{2.}$

Enjoy Maths.!

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

2
Sep 6, 2017

$\frac{1}{2}$

#### Explanation:

Observ that
$\frac{1}{3} = \frac{1}{2} \left(1 - \frac{1}{3}\right)$
$\frac{2}{15} = \frac{1}{2} \left(\frac{1}{3} - \frac{1}{15}\right)$
$\frac{3}{105} = \frac{1}{2} \left(\frac{1}{15} - \frac{1}{105}\right)$
$\ldots .$

In the infinite sum the second term of each item can be simplified with the first term of the following item and the second term goes to 0, so the infinite sum is the first term of the first item = $\frac{1}{2}$

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