# Help with Series: Solving ∞ ∑ (n/2^(n-1)) n=1

• djuiceholder
In summary, the conversation was about a series problem that the person was struggling with. They shared their attempt at solving it and asked for help. Another person then provided hints and eventually a solution involving telescoping series. They also mentioned another way to think about the problem. The conversation ended with the participants discussing the solution and the joy of finding it.
djuiceholder
Help with "series"

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

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

What are you supposed to do with it? I would assume you are supposed to determine if it converges, but unless you say I can't be sure. Also, please show us an attempt at the problem. Then we can help you with where you get stuck.

What an interesting piece of work! I wonder who thought that up. It's hypnotic.
Of course it is easy to find the sum on a spreadsheet . . . approximately, I suppose.
I decided it just HAD to be a telescopic series and played with it for a good long time before I found it. Is that enough of a hint?

I started with the nth term written as the difference 4n over 2^n minus 2n over 2^n and then tried to write the 2nd term in the form of the first term with n replaced by n+1 as is required to collapse a telescopic series. Unfortunately that wasn't quite right and nothing I guessed was either. Finally I resorted to adding x to that first term and subtracting x from the second. To find x, I replaced n with n+1 in the first term and set it equal to the second. It came out very nicely and converges to 4 as the spreadsheet predicted, so I expect it is correct. I'm no mathematician, so this may not be the conventional way of summing this series.

I hope I have not spoiled your fun with this intriguing series!

Oh, a nicer way to do it is to guess that the nth term is an+b over 2^n minus a(n+1)+b over 2^(n+1), where a and b are two parameters to be found. This setup guarantees that the series will be telescoping. Set that expression equal to 2n over 2^n and solve for a and b.

There's another way to think about it. Can you sum the series 1+x+x^2+x^3+...? Now think what happens if you take the derivative and substitute a special value of x.

That's quite a find, Dick! Glad you didn't mention it before I had all that fun finding the telescoping series.

It's just the 'standard' way to do it. Glad you had fun!

## 1. What is the formula for solving the given series?

The formula for solving the given series is ∞ ∑ (n/2^(n-1)).

## 2. How do I determine the value of n in the series?

The value of n in the series can be determined by looking at the subscript of the series, which in this case is n=1. This means that n starts at 1 and increases by 1 for each term in the series.

## 3. How do I find the value of ∞ in the series?

In mathematics, ∞ (infinity) is used to represent a value that is larger than any number. In this series, it represents the infinite number of terms that are being added together.

## 4. Can I use a calculator to solve this series?

Yes, you can use a calculator to solve this series. Simply input the formula and the desired value for n, and the calculator will provide the result.

## 5. What is the significance of 2^(n-1) in the series?

2^(n-1) is the denominator of the series and it is a geometric progression. This means that for each term in the series, the denominator increases exponentially by a factor of 2. This is what causes the series to converge to a finite value instead of diverging to infinity.

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