# Another Question Regarding Power Series

1. Nov 1, 2004

I used Mathematica to confirm that

$$\sum _{n=0} ^{\infty} \left[ \frac{\left( -1 \right) ^n \left( 2n \right) x^{2n+1}}{n! \left( n+1 \right) ! 2^{2n+1}} \right] + \sum _{n=0} ^{\infty} \left[ \frac{\left( -1 \right) ^n \left( 2n \right) \left( 2n+1 \right) x^{2n+1}}{n! \left( n+1 \right) ! 2^{2n+1}} \right] = - \sum _{n=0} ^{\infty} \frac{\left( -1 \right) ^n x^{2n+3}}{n! \left( n+1 \right) ! 2^{2n+1}}$$

Does anyone know how to obtain it directly?

Thank you.

2. Nov 1, 2004

### shmoe

Sure, combine the series on the left and notice that when n=0, the coefficient vanishes to get:

$$\sum _{n=1} ^{\infty} \left[ \frac{\left( -1 \right) ^n \left( 2n \right) (2n+2) x^{2n+1}}{n! \left( n+1 \right) ! 2^{2n+1}} \right]$$

Factor the numerator slightly:

$$\sum _{n=1} ^{\infty} \left[ \frac{\left( -1 \right) ^n \left( n \right) (n+1)2^2 x^{2n+1}}{n! \left( n+1 \right) ! 2^{2n+1}} \right]$$

Do some light cancelling, write our power of x in a funny way and lets pull out a -1 for fun:

$$-\sum _{n=1} ^{\infty} \left[ \frac{\left( -1 \right) ^{n-1} x^{2(n-1)+3}}{(n-1)! \left( (n-1)+1 \right) ! 2^{2(n-1)+1}} \right]$$

You are now a change of index variable away from glory.

Last edited: Nov 1, 2004
3. Nov 1, 2004