What Is the Sum of the Series Ʃ n/(n+1)! from n=1 to Infinity?

[e179285]

Ʃ (from n=1 to ınfınıty ) n /(n+1)!

the questions asks sum of the this series.

I try to write this series as Ʃ 1/(n-1)!.(n+1)but ı couldn't simulate any series.

How can ı do this ?

 

[DryRun]

Hi e179285

\sum_{n=1}^{\infty} \frac{n}{(n+1)!}
Use the ratio test.
u_n = \lim_{n\to\infty}\frac{n}{(n+1)!}=\lim_{n\to\infty}\frac{n}{(n+1)n!}<br /> \\u_{n+1} = \lim_{n\to\infty}\frac{(n+1)}{(n+2)!}=\lim_{n \to \infty} \frac{(n+1)}{(n+2)(n+1)n!}=\lim_{n \to \infty} \frac{1}{(n+2)n!}<br />

 

[e179285]

Hi sharks,

what ı can use ratio test for ?

 

[DryRun]

Do you know the ratio test?

Basically, it goes like this:
\lim_{n \to \infty} \frac{u_{n+1}}{u_n}=LIf L < 1, the series converges. If L > 1, the series diverges, and if L = 1, then the result is inconclusive, meaning, the series may either converge or diverge, and you'll have to use another test to find out.

 

[e179285]

ı know ratio test and ı found it is convergent but the question wants finding sum of the series I don't know how to do this

 

[DryRun]

I misread your post. So, you want to find the limit of the series or the limit of the sequence of partial sums:
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+ \; ...<br /> \\\frac{1}{2}+\frac{2}{6}+\frac{3}{24}+ \; ...<br /> \\\frac{1}{2}+\frac{1}{3}+\frac{1}{8}+ \; ...

 

[Infinitum]

Try applying the Taylor series for derivative of f(x) = \frac{e^x}{x} at x=1.

 

[ehild]

Try applying the Taylor series for derivative of f(x) = \frac{e^x}{x} at x=1.

Is not it rather the derivative of f(x) = \frac{e^x-1}{x} at x=1?

ehild

 

[e179285]

thanks for answers :)

 

[Infinitum]

Is not it rather the derivative of f(x) = \frac{e^x-1}{x} at x=1?

ehild

I believe you can do it both ways. If you use f(x) = \frac{e^x}{x} then you get a constant(not giving the value here, because it's the answer!) plus the series of which the sum needs to be found. Subtracting it both the sides would give the sum.

If you use f(x) = \frac{e^x-1}{x} then that extra term gets subtracted in the expansion, and is compensated by the differentiation on LHS, because the function is framed that way.

 

[DryRun]

I'm curious to know how that function for the Taylor series at x=1 was obtained in the first place? Is there a theorem relating those two specific results?

 

[Infinitum]

I'm curious to know how that function for the Taylor series at x=1 was obtained in the first place? Is there a theorem relating those two specific results?

The Taylor series was just an idea to relate the given series, because you see there are factorial terms in the denominator pretty much laid out the way you need them. There was just a need for manipulation to give you the exact series, according to which the function was decided. A different series sum could be found by possibly using f(x)=sin(x)/x^2, it all depends on the given problem. As far as I know, there is no 'theorem' relating the two.

I think the above sum can be evaluated by yet another method(not involving Taylor series), but I haven't been able to complete the solution yet.

 

[ehild]

I'm curious to know how that function for the Taylor series at x=1 was obtained in the first place? Is there a theorem relating those two specific results?

You know the Taylor series of e^x: The terms have n! in the denominator. e^x=\sum_0 ^{\infty} \frac{x^n}{n!}Integrating it, you get n+1! in the denominators:

\int_0^x{e^udu}=e^x-1=\sum_0 ^{\infty} \frac{x^{n+1}}{n+1!}=x+\sum_1 ^{\infty} \frac{x^{n+1}}{n+1!}
Divide both sides by x:

f(x)=\frac{e^x-1}{x}=1+\sum_1 ^{\infty} \frac{x^{n}}{n+1!}

Take the derivative of both sides:

f&#039;(x)=\sum_1 ^{\infty} \frac{n x^{n-1}}{n+1!}

Substituting x=1, you get the series.

 

[ehild]

I believe you can do it both ways. If you use f(x) = \frac{e^x}{x} then you get a constant(not giving the value here, because it's the answer!) plus the series of which the sum needs to be found. Subtracting it both the sides would give the sum.

The derivative of e^x/x is zero at x=1.

ehild

 

[Infinitum]

The derivative of e^x/x is zero at x=1.

ehild

Yes it is, but,

f&#039;(x)= \frac{-1}{x^2} + 0 + \frac{1}{2!} + \frac{2x}{3!} + \frac{3x^2}{4!}+...

Putting x=1, gives you the answer.

Actually, after your idea I realized you can use Taylor series for any

f(x) = \frac{e^x - a}{x}

at x=1, where a is constant.

 

[ehild]

Yes it is, but,

f&#039;(x)= \frac{-1}{x^2} + 0 + \frac{1}{2!} + \frac{2x}{3!} + \frac{3x^2}{4!}+...

Well, but it is not a Taylor series and not the derivative of a Taylor series.

ehild

 

[tt2348]

A quick way to evaluate this without using Taylor polynomials, e^x, etc is to make it into a telescoping series. n/(n+1)!= n/(n+1)! -1/(n+1)! + 1/(n+1)! = 1/n! -1/(n+1)! ... Plug in the first few terms to see only the first term and the last term remains... Since your last term goes to 0, your first term in your answer

 

[Infinitum]

A quick way to evaluate this without using Taylor polynomials, e^x, etc is to make it into a telescoping series. n/(n+1)!= n/(n+1)! -1/(n+1)! + 1/(n+1)! = 1/n! -1/(n+1)! ... Plug in the first few terms to see only the first term and the last term remains... Since your last term goes to 0, your first term in your answer

Here is the elegant method!

Well, but it is not a Taylor series and not the derivative of a Taylor series.

ehild

As I know it, Taylor series expansion about a point x=a is given by

f(x) = f(a) + f&#039;(a)(x-a) + \frac{f&#039;&#039;(a)(x-a)^2}{2!} + ...

Using that for e^x at a=0, we get the expansion,

e^x = e^0(1+ (x-0) + \frac{(x-0)^2}{2!} + ...)

Which I used in the above post.

 

[ehild]

A quick way to evaluate this without using Taylor polynomials, e^x, etc is to make it into a telescoping series. n/(n+1)!= n/(n+1)! -1/(n+1)! + 1/(n+1)! = 1/n! -1/(n+1)! ... Plug in the first few terms to see only the first term and the last term remains... Since your last term goes to 0, your first term in your answer

Excellent!

ehild

 

[ehild]

As I know it, Taylor series expansion about a point x=a is given by

f(x) = f(a) + f&#039;(a)(x-a) + \frac{f&#039;&#039;(a)(x-a)^2}{2!} + ...

Yes, that is why a series with negative power of x (or x-a) is not a Taylor series .

But your Taylor expansion method was a splendid idea!


ehild

 

[Infinitum]

Yes, that is why a series with negative power of x (or x-a) is not a Taylor series .

Indeed! I should have added that the division by x was separately done, after the Taylor expansion.

 

[DryRun]

A quick way to evaluate this without using Taylor polynomials, e^x, etc is to make it into a telescoping series. n/(n+1)!= n/(n+1)! -1/(n+1)! + 1/(n+1)! = 1/n! -1/(n+1)! ... Plug in the first few terms to see only the first term and the last term remains... Since your last term goes to 0, your first term in your answer

I've been wondering about the possibility of arranging the series into a telescoping series. Glad to know it's possible, so the problem becomes much easier.