Calculate Limit of Series: Step-by-Step Guide

In summary, the conversation is about finding the limit of a given sequence and the individual is seeking help in solving it. They have tried a certain method but were unsure about their results. Another individual suggests a different approach and provides a detailed solution with the correct answer. The conversation ends with the individual thanking for the help and stating that they had made an arithmetic error in their own attempt.
  • #1
esuahcdss12
10
0
hey
I am trying to calculate the limit of :

limn→∞(1/2+3/4+5/8+...+2n−1/2^n)
but I am not sure how to solve it, I thought to calculate 2S and than subtract S, but it did not worked well. I did noticed that the denominator is a geometric serie,but I don't know how to continue. could you help?
 
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  • #2
It should have worked well! Your thinking is good.

You are on a very successful path. Perhaps you just didn't quite interpret your results correctly.

Please demonstrate your efforts.

What does 2S - S look like?
 
  • #3
As suggested, I would continue on your current path, but I wanted to show how I would approach the problem...

I would begin by expressing the partial sum $S_n$ in the following difference equation:

\(\displaystyle S_{n}-S_{n-1}=(2n-1)2^{-n}\)

We see the homogeneous solution is:

\(\displaystyle h_n=c_1\)

And the particular solution will have the form:

\(\displaystyle p_n=(An+B)2^{-n}\)

Substituting $p_n$ into our difference equation, there results:

\(\displaystyle (An+B)2^{-n}-(A(n-1)+B)2^{-(n-1)}=(2n-1)2^{-n}\)

Multiply through by $2^{-n}$:

\(\displaystyle (An+B)-2(A(n-1)+B)=2n-1\)

\(\displaystyle An+B-2(An-A+B)=2n-1\)

\(\displaystyle An+B-2An+2A-2B=2n-1\)

\(\displaystyle -An+2A-B=2n-1\)

Equating like coefficients, we obtain:

\(\displaystyle -A=2\implies A=-2\)

\(\displaystyle 2A-B=-1\implies B=-3\)

And so our particular solution is:

\(\displaystyle p_n=-(2n+3)2^{-n}\)

And hence, the general solution is:

\(\displaystyle S_n=h_n+p_n=c_1-(2n+3)2^{-n}\)

Using the initial value, we find:

\(\displaystyle S_1=c_1-(2+3)2^{-1}=\frac{1}{2}\implies c_1=3\)

And so the solution satisfying all given conditions is:

\(\displaystyle S_n=3-(2n+3)2^{-n}\)

Thus, we find:

\(\displaystyle S_{\infty}=\lim_{n\to\infty}S_n=3\)
 
  • #4
I manged to solve the question, i made an arithmetic error.
thanks
 

Related to Calculate Limit of Series: Step-by-Step Guide

What is a limit of a series?

A limit of a series is the value that the series approaches as the number of terms in the series increases. It is the value that the terms of the series get closer and closer to as the series continues.

Why is it important to calculate the limit of a series?

Calculating the limit of a series can help determine the overall behavior and convergence/divergence of the series. It can also help in solving real-world problems and making predictions based on patterns in the series.

What is the process for calculating the limit of a series?

The process for calculating the limit of a series involves finding the general term of the series, taking the limit of this general term as the number of terms approaches infinity, and then evaluating the resulting expression to find the limit value.

What are some common methods for evaluating the limit of a series?

Some common methods for evaluating the limit of a series include using the ratio test, the root test, and comparison tests. These methods involve comparing the given series to a known series with a known limit value.

Are there any limitations to calculating the limit of a series?

Yes, there are certain series that do not have a well-defined limit, such as oscillating series or series with alternating signs. In these cases, it may not be possible to calculate the limit using traditional methods.

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