Amazing! Okay, how about a "harder" one. Which one is largest/smallest:

I'll leave the proofs to you guys. :-)

Amazing! Okay, how about a "harder" one. Which one is largest/smallest:

I'll leave the proofs to you guys. :-)

Comments temporarily disabled.

New to Spiked Math?

View the top comics.

**New Feature:** Browse the archives in quick view! Choose from a
black,
white or
grey background.

View the top comics.

(Ranked by SM-CRA)

Other Sites:

http://www.wolframalpha.com/input/?i=sum+n%3D0+to+infinity+n%2F2^%28n%2B1%29

Gah - I used to be able to do this stuff when I was starting my engineering major. Now that I've spent a year in Japanese language school, I've forgotten everything...

I'm not very good at math, but I think it's the 1st that's the smallest and the third because the denominator will infinitely get larger while the numerator will stay constant.

The third will stay the largest because it's the only one that it's denominator will be the only one not getting larger after every iteration.

As a side note, sum n=0 to inf of 9/10^(n+1) gives one as well, i.e., 0.999... = 1.

Wow, I thought a little about the proof of the first two being equal (for the case with 2's), and I loved it. The one I thought (I don't know if there are simpler ones) is nice.

As for the series with 3's, I'm not sure if they are equal like in the other case, but I can say that the first one is greater than the third one, since the first term equals 1, and every other term is positive.

In fact, the 1/3^n series equals to $1.50, since it's also a geometric series with a=1 and r=1/3, hence it equals to 1 / (1 - 1/3) = 3/2.

For the homeworkproof: d/dr ar^n=n ar^(n-1). Choosing the value for a correctly will lead to the proof.

Since I thought the proof was extremely elegant I thought I'd share it here:

http://i.imgur.com/AlxOR.png .

It looks fancy (although it's a standard trick for subject that I recently had :P), but generally you need to explain why you are allowed to do what you just did (sort out elemenents the way you like it - commutative and associative property). Just in case you didn't know that, you might like to remember it for the future as it is very important in math analysis. ;)

Same thing for expressions inside the brackets - how do you conclude that the answers of all brackets are similar and follow that pattern. It's easy to do it with induction, formal enough as well.

My solution uses the following:

Sum{n = 0 to inf}(n/2^(n+1)) = Sum{k = 1 to inf}[Sum{n = k to inf}( 1/2^n)]

Mike, I hope I wasn't too harsh with my comment from the survey (I suppose I was the only Croat taking it, so you'll easily find me :P). Still, I love how you changed Spiked Math! Thank you very much and keep up the great work!

Although this task was pretty easy for me (not boasting, I just feel damn proud and thankful of knowing and understanding math a lot :D), I liked reading it and even solving it to check out all the answers! And you also added a nice thought there that made me think a bit (for example, how n/2^(n+1)>1/2^(n+1), but the sum is still equal) Thank you very much. :)

I feel like a proud fan right now, especially when I usually recommend your website to some of my friends, even if they're not mathematicians. :)

Sorry, but you were not the only Croat taking the survey. =P

27 people who took the survey listed Croatia as their country.

Well, that's a shock! Seriously?

...well, great then! I'm damn happy about that! :D

No need to be sorry, in fact, I'm damn glad!

Greetings from one happy tovar. ;)

@Phoenix, I don't see any harsh comments from Croats. However, one did leave the comment "penis" with no other context.

HAHAHA that is hilarious

or maybe I'm not mature enough... one of the two

[I remember taking the survey, but not at all what I wrote, might've left something similar]

I was able to crank out the sum n/3^n by calculating a few terms and guessing well that the sum of the first n-1 terms is 3/4[1 - ( 2n + 1 )/3^n]. Then that can be proved via induction, taking the limit yields the sum is 3/4. Doesn't feel very elegant however.

In the "harder" problem, the first solution yields more than one dollar, because 3^^0=1, and from them you just add positive stuff. So I'll put my bet on it. But reaaaaly I do not want to wait for the clerk to account for all of the fractions in cash...

Another approach to the proof is to expand the sum (1/2 + 1/4 + ...), which we know is 1, square this (so it's still 1), and look at the terms.

For the 3's these series look to me like they converge to 3/2, 3/4, and 3/3.

I hate my intuition. It always tricks me...

First serie = 3/2

Second serie : sum(n/3^n,n,0,inf)

=sum(sum(1/3^t,t,n,inf),n,1,inf)

=sum(1/2*(1/3)^n,n,1,inf)=3/4

I'm having trouble proving that the third number is equal to 1.

Generally it can be concluded that

sum(n=0 to infinity) n/(k^n)=k/(k-1)^2

Do you want proof of this??

I think the proof from the BOOK goes like

k* sum(n≥0) n/k

^{n}= k* sum(n≥1) n/k

^{n}= k* sum(n≥0) (n+1)/k

^{n+1}= sum(n≥0) n/k

^{n}+ sum(n≥0) 1/k^{n}and therefore

sum(n≥0) n/k^n = 1/(k-1) * sum(n≥0) 1/k^

^{n}= k/(k-1)^^{2}for k

an easy proof is:

sum(n/k^n) = k*sum(n/k^(n+1)) = k diff(sum(k^-n)) = k diff(1/(1-k)) = k / (1-k)²

I meant for k

And I am completely stupid...

3rd try:

I meant k < 1

haha

"Afterall, for every n>1, ...."

is misleading since the n=0 terms contributes too much =p

Phoenix said: "And you also added a nice thought there that made me think a bit (for example, how n/2^(n+1)>1/2^(n+1), but the sum is still equal)"

That's because Mike cheated a bit. He states "for every n>1", but the sum starts in "n=0", and that's where our intuition fails us. Both are geometric progressions, but "n/2^(n+1)" has 1/4 as initial term , while "1/2^(n+1)" has 1/2.

By the way, Mike, nice reboot of the page! I definitely like this new organization, and I'm looking forward for new MFT!

Yeah, I figured that out while I was explaining that to someone else. Both numbers are still pretty small since the denominator of each fraction is 2^(n+1) - that's generally a big number!

It was still an interesting aspect, though. :)

For the second series, the first term is 0, but for the first series, the first term is 1/2.

Try this sequence compared to the three existing powers-of-2:

\sum_{n = 1}{\infty} \frac{F_n}{2^{n+1}}

Where F(0) = 0, F(1) = 1, and F(n) = F(n-1) + F(n-2) for n ≥ 2.

the first term of the first sequence is 1/2 while that of the second is 0.

Super Duper easy solution for part proof that S

^{2}= 1:u

^{2}= n*u^{1}u

^{2}≥ u^{1}u

^{2}< 1THEREFORE, 1 ≥ s

^{2}≥ s^{1}lim 1 = 1

lim s

^{1}= 1By the squeeze theorem, 1 ≥ s

^{2}≥ 1Therefore, s

^{2}= 1