Finding the Sum of a Series: A Challenging Problem

[Bob Busby]

Homework Statement

Find the exact sum of the series:

1/1!3 + 1/2!4 + ... + 1/n!(n+2)

Homework Equations

The Attempt at a Solution

The only series I know that look like this are e^x and ln(x) (centered at x = 1). But I do not know how to combine them and I'm not even sure if that's what I'm supposed to do.

e^x = 1 + x + x^2/2! + ... + x^n/n!

ln(x) = (x-1) -(1/2)(x-1)^2 + (1/3)(x-1)^3 ...

ln(x) is missing the factorials and has alternating signs. I tried messing around with its input value but to no avail.

Any help would be appreciated.

 

[Dick]

What kind of a series do you get if you integrate e^x? Can you find a way to modify that function so you get your series?

 

[Bob Busby]

Thanks for the help.

integral(e^x) = x + x^2/2! + x^3/3! + ... + x^(n+1)/n!(n+1)

I had tried that before. Certainly the integral of e^x is the function that most resembles the series, but the values in the denominator of the series' terms are increasing linearly and any thing I plug in for the x's increase exponentially. I thought maybe integrating a second time would do it but that gives the series:

x^(n+1)/n!(n+1)(n+2)

Is there something to multiply the series by to get rid of the (n+1) for all of the terms? That would once again require an input that could increase linearly. I'm kind of lost.

 

[Dick]

Thanks for the help.

integral(e^x) = x + x^2/2! + x^3/3! + ... + x^(n+1)/n!(n+1)

I had tried that before. Certainly the integral of e^x is the function that most resembles the series, but the values in the denominator of the series' terms are increasing linearly and any thing I plug in for the x's increase exponentially. I thought maybe integrating a second time would do it but that gives the series:

x^(n+1)/n!(n+1)(n+2)

Is there something to multiply the series by to get rid of the (n+1) for all of the terms? That would once again require an input that could increase linearly. I'm kind of lost.

How about if you integrate e^x from 0 to 1? And sure that only gives you n+1 in the denominator. How about x*e^x?

 

[Bob Busby]

How about if you integrate e^x from 0 to 1? And sure that only gives you n+1 in the denominator. How about x*e^x?

Thanks again.

I'm not sure what your second sentence means. I'm pretty sure I'll have to stick in a one by integrating from 0 to 1 in the end but I haven't found the right series yet. x*e^x gives me the same thing as the integral of e^x, but with the factorials shifted to the left. I still have the same problem as before; I can't figure out how to make the linearly increasing fractions appear.

 

[Dick]

Thanks again.

I'm not sure what your second sentence means. I'm pretty sure I'll have to stick in a one by integrating from 0 to 1 in the end but I haven't found the right series yet. x*e^x gives me the same thing as the integral of e^x, but with the factorials shifted to the left. I still have the same problem as before; I can't figure out how to make the linearly increasing fractions appear.

Ok, what do you get if you integrate the series for x*e^x term by term from 0 to 1? I'm pretty sure your series is in there.

 

[Bob Busby]

I get the right answer.

Thanks so much for all your help!