SUM of the Series
Will the SUM of the following series ever reach 100 ?
1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 . . .
1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 . . .
Labels: mathemagic, outside-the-box
posted by Rajesh Lal at
10:04 PM
21 Comments:
I'm sure it could, and it would take a LONG time, but I have absolutely nothing to back that up...
no it cant it wont even get to 1
Yes, the sum can get to any number you want.
Take the first term = 1/2
Take the next 2 terms = 1/3 + 1/4, which is more than 1/2
Likewise the next 4 terms = 1/5 + 1/6/+ 1/7 + 1/8 which is also more than 1/2
Similarly for the following 8 terms, then 16 terms and so on, all more than 1/2.
So, whatever number you want to get to, you can reach by accumulating enough terms.
Chris
yes it can..
Your always adding something to it at each itteration, regardless of how small... so with infinite itterations, the value will also be infinite.
just to give you an idea of HOW LONG it will take...
at 1/300000 the sum total of all previous fractions only equals 12.18875509
(Excel is so much fun!!!)
...and by the time you get to 1/1,000,000 you've managed the grand sum of 13.39272672. YAY!
I got somewhere in the ballpark at 1/4.91814569773648*10^43 the sum is approximately 100 but my calculator explodes after that.
of course , accumlating numbers as long as they arnt negative will always reach a certain number, ( only applies to positive) just take a long time
To the multiple people, that think as long as you add something, that is greater than zero infinitely, you can reach any number:
This is incorrect. For example, 1+1/2+1/4+1/8+... = 2.
This problem's series is known as the harmonic series (but you need to add '1+' to the front). It is divergent, and the sum of its first n terms is around ln(n), where ln is the natural logarithm. So to reach 100, you need about e^100 terms, where e is 2.718281828459045...
Which is around 10^43, as pointed out earlier.
By the way, an easy way to remember the value for e is this little ditty:
"Jackson, Jackson wore a pair of 45's until he was ninety".
Andrew Jackson served 2 (first digit) terms as the 7(second digit)th president of the US. He was elected in 1828 (hence the next 8 digits from 'Jackson, Jackson'), and the two guns and the age give the last six digits. Why is the 90 between the two 45s, you may ask? Because surely he did not wear the two 45's on one hip next to each other!
This sounds a lot like e from calulus
No never, it will never get to one
it will get to one hundred eventually... the first three fractions add up to 13/12. this means the progression is gaining in size... it will take a long time, but it will happen.
whoever said it won't get to one is dumb
Using calculus... this is basically the limit of the sum of 1/N as N goes from 1 to infinity... this limit IS inifnity... it will surpass 100 for sure
If the problem was 1/2 + 1/4 + 1/9 then no, this would never surpass 100
i think we can all rest and agree to the fact it will reach 100.
im not a mathematician so i cant prove this but from common sense its possible.
its infinity which is the limit as pointed out. u can add fractions million, billion or trillion times it will reach there.
and i alsoo think it could reach 1,000,000 and above if you kept on going...doesnt need to stop at 100.
Huzzy
http://www.math.unh.edu/~jjp/proof/proof_n.html
harmonic series diverges, even if we throw out a finite number of terms it still diverges, if you throw out a infinite number then you can find a sub-sequence that converges (geometric series sum(2^-n) = 2)
really more real analysis than calculus
To Sum Is To Add
in theory it cant with that calculus stuff already talked about here. but in reality, with an infinte amount of this series, it will eventually surpass the number because you are forever adding something positive.
Will the sum ever eqaul exactly 100? That might be the real question. LOL
The limit approaches infinity since it is a (divergent) harmonic series. The first term does not affect the limit.
Well, it does go past 100, indeed the harmonic series of the form,
H = 1 + 1/2 + 1/3 + .... + 1/n + ... ad inf.
is divergent, so its partial sums are unbounded.
However the harmonic numbers that are the partial sums in this case,
H_n = 1 + 1/2 + 1/3 + .... + 1/n
will never take an integer value for any n.
Read,
http://mathrants.blogspot.com
for more!
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