Prove: Sum of Series Equals e^2 - e

[lionely]

Homework Statement

Prove the following result:
\frac{1}{1!} + \frac{1+2}{2!} + \frac {1+2+2^2}{3!} ... = e² - e

Homework Equations

The Attempt at a Solution

Could someone please give me a hint on what to do . I tried writing out the maclaurin series for e^2 and e but don't know what to do with them.

(also why won't the latex work for me?)

 

[Mentallic]

Homework Statement

Prove the following result:
\frac{1}{1!} + \frac{1+2}{2!} + \frac {1+2+2^2}{3!} ... = e² - e

Homework Equations

The Attempt at a Solution

Could someone please give me a hint on what to do . I tried writing out the maclaurin series for e^2 and e but don't know what to do with them.

(also why won't the latex work for me?)

The numerators are a geometric sum.

 

[Curious3141]

Homework Statement

Prove the following result:
\frac{1}{1!} + \frac{1+2}{2!} + \frac {1+2+2^2}{3!} ... = e² - e

Homework Equations

The Attempt at a Solution

Could someone please give me a hint on what to do . I tried writing out the maclaurin series for e^2 and e but don't know what to do with them.

(also why won't the latex work for me?)

Mentallic has already provided a hint - find an expression for the $k$th term and put it in summation notation to make it clearer.

And to format in LaTex, you have to enclose the text within tex tags like this (quote my post to see):

\frac{1}{1!} + \frac{1+2}{2!} + \frac {1+2+2^2}{3!} ... = e^2 - e

or you can use the dollar sign shorthand notation like so:


$$\frac{1}{1!} + \frac{1+2}{2!} + \frac {1+2+2^2}{3!} ... = e^2 - e$$

 

[lionely]

Is the nth term, [1 + 2^(n-2) + 2^(n-1)]/n! ?

 

[Mark44]

(also why won't the latex work for me?)

See https://www.physicsforums.com/showpost.php?p=3977517&postcount=3, especially the section titled Simplified tags.

 

[Curious3141]

Is the nth term, [1 + 2^(n-2) + 2^(n-1)]/n! ?

Seems to work for n=3 but why don't you try other values of n?

Mentallic already gave you the pattern for the numerator. How do you find the sum of that pattern?

 

[Mentallic]

Is the nth term, [1 + 2^(n-2) + 2^(n-1)]/n! ?

No, the n^th term is

\frac{1+2+2^2+2^3+...+2^{n-1}}{n!} = \frac{2^0+2^1+2^2+2^3+...+2^{n-1}}{n!}

 

[HallsofIvy]

No, the n^th term is

\frac{1+2+2^2+2^3+...+2^{n-1}}{n!} = \frac{2^0+2^1+2^2+2^3+...+2^{n-1}}{n!}

= \frac{2^{n+1}- 1}{n!}

lionely did you not notice that the numerators are 1, 1+ 2= 3, 1+ 2+ 4= 7, 1+ 2+ 4+ 8= 15, 1+ 2+ 4+ 8+ 16= 31, etc., all one less than a power of two?

 

[Mentallic]

= \frac{2^{n+1}- 1}{n!}

Actually,

\frac{2^n-1}{n!}

 

[lionely]

Would this be proving it , if I find the series for e^2 and e^1 and just add them? Because with the nth term I don't see how to show that is e^2 -e

 

[Mentallic]

Can you apply the geometric sum formula for each numerator?

 

[lionely]

What would the common ratio be , 2?

 

[Mentallic]

Yes. You should brush up on geometric sums because by this point it's assumed that you can spot them and apply the rules correctly without hassle.

 

[lionely]

I'm sorry but I'm still confused, I tried grouping the terms that are powers of two, but all the denominators are different. So how would I go about finding the sum to infinity of that?

 

[Mentallic]

Show me what you have so far.

 

[lionely]

so the series is (1/1!) + (1+2/2!) + (1+2+2^2/3!) ... right?

so I tried grouping the 2's like this

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

 

[Mentallic]

No, you're going about this all wrong.

We have

\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+2^2}{2!}+\frac{1+2+2^2+2^3}{3!}+...+\frac{1+2+2^2+...+2^{n-1}}{n!}+...

Now we need to apply the geometric sum formula to each numerator. 1+2+2² can be simplified using the formula. 1+2+2²+2³ can be simplified. etc.

What is 1+2+2² equivalent to using the formula?

 

[lionely]

1/(1-2) ? also I just remembered just this only work if |r|<1?

 

[Mentallic]

1/(1-2) ? also I just remembered just this only work if |r|<1?

That obviously can't be right. Does

1+2+2^2=\frac{1}{1-2}

?

You're thinking of infinite geometric summations, while we are dealing with finite sums. Just google geometric sum, read a little about it and it should all come back to you.

 

[lionely]

Oh geometric sum, I kept thinking about summing to infinity. Um it would be (2^3-1)/2-1

 

[lionely]

Okay so basically we end up with (1/1!) + (3/2!) + (7/3!) + (15/4!) ... (2^[n-1]/n!)

But I'm sorry I still can't see how this will shape up to be e^2 -e! :(

 

[Mentallic]

Ok good, so can you now simplify each numerator in the expression

\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+2^2}{2!}+\frac{1+2+2^2+2^3}{3!}+...+\frac{1+2+2^2+...+2^{n-1}}{n!}+...

edit: Leave the numerator in terms of the result you get from the geometric sum.

Since

1+2+2^2+...+2^{n-1}=2^n-1

by the geometric sum formula, we have the expression above equal to

\frac{2^1-1}{1!}+\frac{2^2-1}{2!}+\frac{2^3-1}{3!}+...+\frac{2^n-1}{n!}+...

Does this make sense? Can you figure out what to do now?

 

[lionely]

Oh I see, I'm an idiot. Thanks for helpign me realize, sigh sigh sigh... sigh. the 2 are e^2 and the -1s are e...

therefore the Sum to Infinity is e^2 -e. Thanks for your help, I really appreciate it.

 

[Mentallic]

You're still missing something though.

e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}=\frac{x^0}{0!}+\frac{x^1}{1!}+...=1+\frac{x}{1!}+\frac{x^2}{2!}+...

Notice the 1 value at the start of the sum. Since the expression you were given doesn't have that, we will have

\left(\frac{2^1}{1!}+\frac{2^2}{2!}+...\right)-\left(\frac{1}{1!}+\frac{1}{2!}+...\right)=\left(e^2-1\right)-\left(e-1\right)=e^2-e

You'd lose marks for jumping straight to the answer or at least not realizing that it wasn't the entire Maclaurin series.