- #1

- 129

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[itex]\sum_{i=1}^\infty \frac{i}{2^i}[/itex]

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- Thread starter Max.Planck
- Start date

- #1

- 129

- 0

[itex]\sum_{i=1}^\infty \frac{i}{2^i}[/itex]

- #2

- 22,129

- 3,297

Do you know the sum of

[tex]\sum_{n=1}^{+\infty} x^n[/tex]

for |x|<1 ??

Now take derivatives.

[tex]\sum_{n=1}^{+\infty} x^n[/tex]

for |x|<1 ??

Now take derivatives.

- #3

- 129

- 0

Do you know the sum of

[tex]\sum_{n=1}^{+\infty} x^n[/tex]

for |x|<1 ??

Now take derivatives.

I know the sum, it evaluates to:

[tex]\frac{1}{1-x}-1[/tex]

But, how will the derivatives help? The i-th derivative with respect to x gives me:

[tex]\sum_{n=1}^{\infty} \frac{n!}{(n-i)!}x^{n-i}[/tex]

Last edited:

- #4

- 606

- 1

I know the sum, it evaluates to:

[tex]\frac{1}{1-x}-1[/tex]

But, how will the derivatives help? The i-th derivative with respect to x gives me:

[tex]\sum_{n=1}^{\infty} \frac{n!}{(n-i)!}x^{n-i}[/tex]

Why would you take the i-th derivative if you were said "

[tex]\frac{1}{1-x}-1=\sum_{n=1}^\infty x^n\Longrightarrow \frac{1}{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}[/tex]

and now just choose a convenient x within the convergence radius...

DonAntonio

- #5

- 296

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The nth term of the sum can be written as

[tex]\left(\frac{1}{2^n}+\frac{1}{2^n}...+\frac{1}{2^n}\right)[/tex]

where we have the fraction repeated n times. Now look at this structure:

1/2

1/4+1/4

1/8+1/8+1/8

.............

Now sum columns instead of the rows!

- #6

- 129

- 0

Why would you take the i-th derivative if you were said "the derivative? I find that i as index pretty confussing and annoying, so I'll change it:

[tex]\frac{1}{1-x}-1=\sum_{n=1}^\infty x^n\Longrightarrow \frac{1}{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}[/tex]

and now just choose a convenient x within the convergence radius...

DonAntonio

But how do you choose that x? Shouldn't

[tex]x^{n-1}=(1/2)^{n}[/tex]

But then x will be a term dependent on n...

- #7

- 606

- 1

But how do you choose that x? Shouldn't

[tex]x^{n-1}=(1/2)^{n}[/tex]

But then x will be a term dependent on n...

...and thus [itex]\,\displaystyle{ x^{n-1}=\frac{1}{2}\left(\frac{1}{2}\right)^{n-1}}[/itex]...! And of course, taking off a factor of 0.5, x is a constant.

DonAntonio

- #8

- 129

- 0

...and thus [itex]\,\displaystyle{ x^{n-1}=\frac{1}{2}\left(\frac{1}{2}\right)^{n-1}}[/itex]...! And of course, taking off a factor of 0.5, x is a constant.

DonAntonio

Ah right lol, thank you very much.

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