What is the summation of this partial sequence?

[Amaelle]

Homework Statement

look at the image

Relevant Equations

Sommation
taylor series

Greetings!
I want to caluculate the summation of this following serie

I started by removing the 4 by

and then

and I thought of the taylor expansion of
Log(1-x)=-∑xⁿ/n but as the 2 is not inside (-1,1) I couldn´t use it
any hint?
thank you!
Best !

 

[BvU]

I started by removing the 4

That is derailing spectacularly ! Write down a few terms of the one and the other and see that the two are very different !
Also: renaming an index ($n$ here) works better if you use a different symbol (e.g. use $k = n+4$) and that changes the summation limits to:
$n=0,\quad k = n +4 \Rightarrow$ lower limit is $k = 4$
$n=N,\quad k = n+4 \Rightarrow$ upper limit is $k = N+4$

And in the numerator you have $n=k-4$ so the exponent is not $k$ but $k-4$

$\$

 

[Amaelle]

Homework Statement:: look at the image
Relevant Equations:: Sommation
taylor series

Greetings!
I want to caluculate the summation of this following serie
View attachment 297263
I started by removing the 4 by
View attachment 297264
and then

View attachment 297265
and I thought of the taylor expansion of
Log(1-x)=-∑xⁿ/n but as the 2 is not inside (-1,1) I couldn´t use it
any hint?
thank you!
Best !

you are right thank you
and then? the main problem remain unsolved?

 

[anuttarasammyak]

If you have already studied calculus, you may make use of the relation
$$\int_{-\infty}^1 2^{mt} dt = [\frac{2^{mt}}{m \ln 2}]_{-\infty}^1=\frac{1}{\ln 2}\frac{2^m}{m}$$
for easy summation.

 

[Amaelle]

If you have already studied calculus, you may make use of the relation
$$\int_{-\infty}^1 2^{mt} dt = [\frac{2^{mt}}{m \ln 2}]_{-\infty}^1=\frac{1}{\ln 2}\frac{2^m}{m}$$
for easy summation.

thank you!

 

[BvU]

the main problem remain unsolved

Haha, if you're starting in the wrong direction it won't be solved anyway

I don't have all the answers but a modest beginning might come about by writing down a bunch of terms for a low value of $N$ ...

Homework Statement:: look at the image
Relevant Equations:: Sommation
taylor series

I want to caluculate

So the actual homework statement is
Homework Statement:: 'Calculate (or: find a compact expression for) $\quad \displaystyle \sum_{n=0}^N \ {2^n\over n+4}\quad$ '

[edit] Ah, we have another contributor !

$\$

 

[Amaelle]

Haha, if you're starting in the wrong direction it won't be solved anyway

I don't have all the answers but a modest beginning might come about by writing down a bunch of terms for a low value of $N$ ...Indeed, that was a huge contribution! I can do this kind of deadly mistakes during exams, thanks a million for point it out!

 

[Fred Wright]

I doubt you will see this problem on an exam. The solution is rather difficult resulting in a complex valued function involving the Lerch transcendent. I'm surprised that it was issued as a homework problem (that's just mean).

 

[Amaelle]

I doubt you will see this problem on an exam. The solution is rather difficult resulting in a complex valued function involving the Lerch transcendent. I'm surprised that it was issued as a homework problem (that's just mean).

Indeed, my instructor said he can asked it in the exam, so I tried to solved it, :)

 

[vela]

I thought of the taylor expansion of
Log(1-x)=-∑xⁿ/n but as the 2 is not inside (-1,1) I couldn´t use it

Besides convergence of the Taylor series, you're trying to calculate a finite sum whereas the Taylor series is an infinite sum.

Here's a hint:
$$\sum_{k=4}^{N+4} \frac{x^k}{k} = \sum_{k=4}^{N+4} \int x^{k-1}\,dx$$ Remember that you're evaluating a finite sum, so the problem actually isn't too bad.