problem of power series

**somnath6088** · July 4th 2008, 08:26 PM

1)Prove that
1)1+ 1/2! + 1.3/4! + 1.3.5/6! + ……………………=e^1/2
2) 2) Evaluate the series
1 + 3/1! + 5/2! + 7/3! + …………………………….

**mr fantastic** · July 4th 2008, 09:00 PM

Originally Posted by somnath6088

1)Prove that
1)1+ 1/2! + 1.3/4! + 1.3.5/6! + ……………………=e^1/2
2) 2) Evaluate the series
1 + 3/1! + 5/2! + 7/3! + …………………………….

Both problems rely on you using the power series $e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + .... + \frac{x^n}{n!} + ..... = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ .

1) Substitute $x = \frac{1}{2}$ and play around a little bit.

2) The series can be written $\sum_{n=0}^{\infty} \frac{(2n+1)}{n!} = 2 \sum_{n=1}^{\infty} \frac{1}{(n-1)!} + \sum_{n=0}^{\infty} \frac{1}{n!} = 2 \sum_{m=0}^{\infty} \frac{1}{m!} + \sum_{n=0}^{\infty} \frac{1}{n!} = 2 e^1 + e^1 = 3 e$ .

**Soroban** · July 5th 2008, 05:29 AM

Hello, somnath6088!

mr fantastic has the best approach to #1 . . .
. . I did it head-on.
We know that: . $e^x\;=\;1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \hdots$

Prove that: . $1+ \frac{1}{2!} + \frac{1\cdot3}{4!} + \frac{1\cdot3\cdot5}{6!} + \frac{1\cdot3\cdot5\cdot7}{8!}+ \hdots \:=\:e^{\frac{1}{2}}$

Reduce the fractions . . .

$S \;=\;1 + \frac{1}{2} + \frac{1}{2\cdot4} + \frac{1}{2\cdot4\cdot6} + \frac{1}{2\cdot4\cdot6\cdot8} + \hdots$

. . $= \;1 + \frac{1}{2(1)} + \frac{1}{2^2(1\cdot2)} + \frac{1}{2^3(1\cdot2\cdot3)} + \frac{1}{2^4(1\cdot2\cdot3\cdot4)} + \hdots$

. . $= \;1 + \frac{\frac{1}{2}}{1!} + \frac{(\frac{1}{2})^2}{2!} + \frac{(\frac{1}{2})^3}{3!} + \frac{(\frac{1}{2})^4}{4!} + \hdots$

And this is the infinite series for: . $e^{\frac{1}{2}}$

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