What is the summation of this partial sequence?

In summary: Besides convergence of the Taylor series, you're trying to calculate a finite sum whereas the Taylor series is an infinite sum.$$\sum_{k=4}^{N+4} \frac{x^k}{k} = \sum_{k=4}^{N+4} \int x^{k-1}\,dx$$$$\sum_{k=4}^{N+4} \frac{x^k}{k} = \sum_{k=4}^{N+4} \int x^{k-1}\,dx$$The Taylor series for ##x## is:$$\displaystyle \sum_{n=0}^N \ {2^n
  • #1
Amaelle
310
54
Homework Statement
look at the image
Relevant Equations
Sommation
taylor series
Greetings!
I want to caluculate the summation of this following serie
CodeCogsEqn.gif
I started by removing the 4 by
CodeCogsEqn (1).gif

and then

CodeCogsEqn (2).gif

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 !
 
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  • #2
Amaelle said:
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##

##\ ##
 
  • #3
Amaelle said:
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?
 
  • #4
If you have already studied calculus, you may make use of the relation
[tex] \int_{-\infty}^1 2^{mt} dt = [\frac{2^{mt}}{m \ln 2}]_{-\infty}^1=\frac{1}{\ln 2}\frac{2^m}{m}[/tex]
for easy summation.
 
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  • #5
anuttarasammyak said:
If you have already studied calculus, you may make use of the relation
[tex] \int_{-\infty}^1 2^{mt} dt = [\frac{2^{mt}}{m \ln 2}]_{-\infty}^1=\frac{1}{\ln 2}\frac{2^m}{m}[/tex]
for easy summation.
thank you!
 
  • #6
Amaelle said:
the main problem remain unsolved
Haha, if you're starting in the wrong direction it won't be solved anyway :smile:

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## ...

Amaelle said:
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 !

##\ ##
 
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  • #7
BvU said:
Haha, if you're starting in the wrong direction it won't be solved anyway :smile:

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!
 
  • #8
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).
 
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  • #9
Fred Wright said:
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, :)
 
  • #10
Amaelle said:
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.
 
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Related to What is the summation of this partial sequence?

1. What is a partial sequence?

A partial sequence is a sequence of numbers that is only a portion of a larger sequence. It may be a finite or infinite sequence, but it does not include all the terms of the larger sequence.

2. What is summation?

Summation is a mathematical operation that involves adding together a sequence of numbers. It is often represented by the symbol ∑ and is used to find the total value of a series of numbers.

3. How do you calculate the summation of a partial sequence?

To calculate the summation of a partial sequence, you simply add together all the numbers in the sequence. If the sequence is finite, you can do this manually. If the sequence is infinite, you can use a mathematical formula or a computer program to calculate the summation.

4. Why is the summation of a partial sequence important?

The summation of a partial sequence is important because it allows us to find the total value of a sequence of numbers without having to add them individually. It is a useful tool in many areas of mathematics and science, such as in calculating probabilities, finding areas under curves, and solving differential equations.

5. Can the summation of a partial sequence be negative?

Yes, the summation of a partial sequence can be negative. This can happen if the sequence contains both positive and negative numbers, and the negative numbers have a greater magnitude than the positive numbers. In this case, the summation would result in a negative value.

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