In maths what does the "sum to infinity" mean?
928453
It means you keep summing forever, that is to say the series converges to a value at infinity.
951210
ok...well 1/x becomes basically infinity as x gets really close to zero...so (1/x)/x woud be infinity/something really close to 0 which would make the value infinity\n\nso we have\n\ne-(1-x)^infinity\nwell...thats ok\n\ne-(1 - (basically zero))^infinity is\ne-1^infinity\nwhich is\ne-1\n\nso the limit as x approachs zero from the right is e-1 which is approximately 1.71828\n\nhope this helps\nmatttlocke
1136716
2r<22\nr<11\nAnswer= (-infinity,11)\n\n* don't have the infinity symbol on computer
750000
The Riemann sum is a way of computing integrals by breaking down a curve into small area elements, approximating the area of each element, adding them together, then taking the limit as the number of elements goes to infinity.  You can find a good discussion with examples here http://en.wikipedia.org/wiki/Riemann_sum
529993
If (1+2+4+8+16+...) is an infinite series you are basically saying that (infinity = 1 + 2*infinity)  which in itself is true because (infinity = infinity + 1) and (infinity = 2 * infinity).  So what you have basically proven is that (S = infinity = infinity) You can't subtract infinity from itself and obtain 1.  You get 0 which means both sides of the equation are equal.\n\nIf you make the series finite, then you can see how the logic doesn't hold to determine that S=-1.\nS = 1+2+4+8+16+32\nS = 1+2(1+2+4+8+16)\nS = 1+2(S-32)\nS = 1+2S-64\n-S = -63\nS = 63
677147
It's a divergent sum for all values of s less than or equal to 1 (and then some other numbers, but I don't know the exact upper bound).\n\n\n\n1+1/(2^s)+1/(3^s)+1/(4^s)+ . . . =1+2^(-s)+3^(-s)+4^(-s)+. . .  is greater than or equal to 1+1/2+1/3+1/4+. . . for s less than or equal to 1.\n\ntherefore, for s=-2, it would diverge.
13310
It's the limit of a type of convergent infinite sum (that's why it's symbol is an elongated 's').\n\nIt's applications range from finding the solutions to differential equations, being used to calculate an area, volume (in the sense we know it, and in a generalised space of n-dimensions).  In Fourier Analysis it's used to approximate functions using cos and sin, etc.  But, by far, the best use of it has to be answering exam questions.\n\nNot helpful, but fairly interesting.  So I thought.  I'll go now.
136287
mathsforums is absolutely correct.  Why are those forms indeterminate?  Because they may be more than one answer!  But you have already pointed them out!  It could be e, 1, or simply there is no limit (diverges).  \n\nThis is fundamental because we are taking limits.  The "1" in 1^infty is not actually the value 1.  It is a limit of a sequence that approaches 1.  If it is actually the value 1, then we can conclude that limit as f(x)->infity 1^f(x) = 1.
313965
This approaches infinity as n approaches infinity.  To demonstrate, I'll show you something even smaller that sums to infinity.\n\n1 + 1/2 + 1/4 + 1/4 + 1/8 [4 times] + 1/16 [8 times]...\n\nThis is equal to 1 + 1/2 + 1/2 + 1/2 + 1/2.....which "goes to infinity.  \n\nSince each term my this series is equal to or smaller then your series (1/4 + 1/4 < 1/3 + 1/4, and 1/8 + 1/8 + 1/8 + 1/8 < 1/5 + 1/6 + 1/7 + 1/8, and so on), and my series approches infinity, your series must approach infinity, too!
1115436
The number is not defined.\n\nLots of people are saying things like...\n"0.000...00001"\n"1/x as x approaches infinity"\n\nThese numbers, technically, are the same as 0.\n\n"n^0" --- ok that's just not thinking. If n <> 0 or n <> inf, then any n^0 = 1.\n\nProof:\nLet 0* be a number infinitely close to 0.\n0* = 1 - .999....\n.999... = 3 * (.333...) // Because 3*3 = 9 no matter how many time you do it.\n.999... = 3 * (1/3 )  // Exact fractional representation\n.999... = 1\nTherefore, 0* = 1 - 1, and 0* = 0.\n\nAs for the answer, the "smallest", non-zero real number does not exist. Well, we know that it "should" exist. It just doesn't.\n\nExcept on computers. That's another problem entirely. Because computers have a data limit, they do actually have a smallest real number >0
957362
we are taught in university that INFINITY is NOT a number, its' a concept. you cant convege to infinity, only diverge because you can never get there. there is no such thing as the highest number....only high numbers
306774
Anything divided by 0 is actually undefined, not infinity or 0. When you think about how that works, it makes sense:\n\nDividing a numerator by smaller numbers makes your number larger. Divide 4 by 2, you get 2. Divide 4 by 1, you get 4. Divide 4 by 0.5, you get 8. And so on until you are dividing by a number so small it's pretty close to 0. The overall number gets larger and goes off into infinity.\n\nBut, if you divide a negative number in the same manner, your number gets smaller. -4 by 2 is -2, -4 by 1 is -4, -4 by 0.5 is -8, etc. As you approach 0, your numbers go off into negative infinity.\n\nYou'll have the same weird effect if you divide by negative numbers - a positive divided by an increasingly smaller negative will yield negative infinity, and a negative divided by an increasingly smaller negative will yield positive infinity.\n\nYou can't have it both ways - dividing by zero yields both a positive and negative infinity! So someone a long time ago decided that since a number can't really have two values, it will be undefined.
494525
Think of it in terms of logarithms, since it is the inverse of exponents.\n\nStatement to prove:  1^infinity = 1\n\nRewritten in logarthmic form:  log[1](1) = infinity\n(Statement above is read: log base 1 of 1 equals infinity)\n\nWell that can't make sense, right?  It is a law of logs and of exponents that log[b](1) = 0 for any positive base b.  This is the same as saying b^0 = 1, for any b.  Also, log[b](b) = 1, for any positive b.  This makes since in b^1 = b.  Both of these laws would conflict this reasoning.  In fact, when b = 1, they conflict with one another.  Therefore, lets look at it with a variable instead.\n\nProve:  1^infinity = X\n\nRewrite in logartithmic form: log[1](X) = infinity\n\nNow, what value of X will give you infinity?\nCan you determine that?\n\nI would say it is indeterminate :)
52450
There is no answer.\n\nWhat you are expressing is the sum of ((-1)^n+1 * 2) from 0 to infinity. This is a divergent series and therefor has no answer, or in mathematica terms, the answer DNE (does not exist).
607462
Its infinity of course
892605
(constant)/(2^n) when n->infinity => the limit is 0(zero)\n\nPS\nNot to be a a stickler for rules but is this really laptop related ? :))
634042
Theres a poster in my maths class that says "a trillion steps towards an infinitae pi" So i assume the best so far is a trillion number after the decimal point. WOW!