How Do You Solve the Series Sum n/2^(n-1)?

In summary, the conversation was about a problem involving an infinite series, specifically ∑ n/2^(n-1) where n=1. The person had learned how to solve different series problems but struggled with this one, spending hours on it. They attempted to simplify it to 2n/(2^n) and considered using the ratio test, but were unsure of how to proceed. Another person suggested using the series 1+x+x^2+x^3+... and taking a derivative. The OP was supposed to evaluate the series but was unsure how to do so. The conversation ended with the OP removing the problem from the original forum they posted it in.
  • #1
djuiceholder
8
0

Homework Statement


so I have learned how to do different problems of series.
but there's this problem that I spent hours last night but could not come up with anything.
which is-


n/2^(n-1)
n=1



Homework Equations



I have no idea

The Attempt at a Solution


so from there i was able to take out 2 and end up with: 2 n /(2^n)

i don't know how to do after that. however, i thought about doing (1/2)^n but then again it does not make sense to me.

help needed. i will appreciate that. thanks
 
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  • #2
Why have you cross-posted this problem to this forum and the Precalc forum?
 
  • #3
Something tells me that the ratio test could helo here.
 
  • #4
If you want to show the series converges, the ratio test can help. If you want to find the sum, the ratio test can't help. Hint: this looks like it's related to the series 1+x+x^2+x^3+... Take a derivative. Now figure out what to put x equal to.
 
  • #5
i was supposed to evaluate it. I don't know how though
 
  • #6
i posted this problem on a different folder and i was warned because it was a wrong place to post. anyways, i think i took it off
 

FAQ: How Do You Solve the Series Sum n/2^(n-1)?

What is an infinite series?

An infinite series is a sum of an infinite number of terms. It can be represented in the form of ∑n=1 an, where an is each term in the series and n represents the index of the term.

What is the formula for the infinite series n/2^n-1?

The formula for the infinite series n/2^n-1 is ∑n=1 n/2^n-1 = 1 + 1/3 + 1/7 + 1/15 + ...

How does the value of n/2^n-1 change as n increases?

As n increases, the value of n/2^n-1 decreases. This is because the denominator, 2^n-1, increases at a faster rate than the numerator, n.

Is the infinite series n/2^n-1 convergent or divergent?

The infinite series n/2^n-1 is convergent. It has a limit of 2, meaning that as n approaches infinity, the sum of the series approaches 2.

How is the convergence of an infinite series determined?

The convergence of an infinite series is determined by finding its limit. If the limit exists and is a finite number, then the series is convergent. Otherwise, it is divergent.

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