does the infinite series defined by the natural log of n devided by n converge?
831566
log(n)/n > 1/n if n>2, and the sum of 1/n diverges (this is the so-called harmonic series), so log(n)/n also diverges.
496208
log(x) - log(1 + (x^2 + 1)^(1/2))\n\nWhere log is the natural logarithm, of course.
602536
Don't believe your have a closed form of U(n). Perhaps, that's because you couldn't type the formula correctly.\n\nOnce you have the correct closed form of U(n), just try to see if you could find the limit as n approaches infinity, then you know whether the sequence converges or not.
823338
That was my answer you quoted above, and when I encountered the "log of a negative number," I did <not> just make a willy-nilly guess, so in my opinion, your characterization was unwarranted.  (In fact, you had asked me to look at this problem in the first place.)\n\nWhen I removed the negative sign, I obtained an answer, and, if you looked at the "sources" section, you'd have seen that it was the right answer.\n\nIn light of your denigration of what you called my "guessing method," I'll add the following (which I felt I didn't have to research in light of the correct answer I obtained and checked) ...\n\nThe logarithms could just as easily have been natural or Napierian logs instead of common logs.  Instead of\nn = (log 0.05) / (log -0.6478), I could have written\nn = (ln 0.05) / (ln -0.6478).\n\nBut ln(-0.6478) = (ln -1) + (ln 0.6478)\nwhich begs the question, what is ln(-1)?\n\nTo answer this, we turn to Euler's famous identity,\n\ne^(i pi) + 1 = 0\ne^(i pi) = -1\n(i pi) ln e = ln(-1)\n\nBut since ln e = 1, we have\n\nln(-1) = i pi\n\nwhich is entirely in the imaginary portion of the complex plane, with no real component.\n\nTherefore, the real part of\n\nln(-0.6478) = (ln -1) + (ln 0.6478) = ln 0.6478\n\nConsidering what I'd done earlier with your question, including checking the series solution in the "sources" section, I thought I'd done enough.  Your characterization of my work is an insult.\n\nThe other answer given here -- a good one -- looks more general to me.  Instead of being based on the Euler identity that I used here, it's based on Euler's more general formula\n\ne^(ix) = cos x + i sin x\n\nNevertheless, I felt it necessary to defend what I'd done.
574891
im pretty sure that you will have to use the parts method to solve it. but the answer is\n\nx(log(1-ie^(ix))-log(1+ie^(ix))+i(Li(-ie^(ix)-Li(ie^(ix))\n\ni=imaginary unit\nlog=natural logarithm\nLi=polylogarithm\n\nyou can check my answer to see if its right because i also thought that it was wrong. i dont know how the imaginary units could have shown up but o well.
695801
I forgot too much about this stuff to prove convergence.  However, if you can take care of that side of things...\n\nKnowing that the sequence converges, then consecutive terms are nearly equal as n grows larger.  This means\n\nlim(n to inf)  X_(n+1) = X_n\n\n0.5(a+X_n) = X_n\n\nX_n = a.\n\nEDIT: I got something better, you can show that\n\nX_n = a/2 + a/4 + a/8 + ... + a/2^n + X_0/2^n\n\nwhere X_0 is the initial term.  Of course X_0/2^n converges to zero so you're left with a geometric series.
1187309
I think I did it.  If y=1/log(x), then a few manipulations gave me\n\nx=10^(1/y).\n\nI'm assuming by log you mean log to the base 10.  I suppose mathematicians will start fussing about what happens at y=0, but a physicist like me doesn't worry that much about infinities ...
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 :)
909632
well the series is only convergent if the partial sums approach zero as n approaches infinity\n\nn^2(e^(-n^3) = n^2/(e^(n^3)\n\nnow take  lim n->inf n^2/(e^(n^3)\n\ne is being rasied to a number that gets very large very quickly as opposed to n squared so this should approach zero\n\nto prove that we can just use L'Hopistal's rule once and see that this then becomes lim n->inf 2/(3ne^(n^3)) this is obviously zero\ntherefore since the partial somes become zero eventually.. the total series converges ... you continue to add zero to the total sum as n goes to infinity\n\nrusty on my calc two hopefuly that helps
1043703
The ... implies an endless continuation of the series.  So the answer is \n\nINFINITY.
576971
If you want to do and infinite sum\nsum_1^infty a_n,\nit is usually done by taking the partial sums\ns_n=a_1+...a_n\nand taking the limit of these partial sums.\nFor Cesaro summability, let\nC_n be the average of the first n partial sums\nC_n=(s_1+...s_n)/n\nand take the limit of C_n as n goes to infinity.\n\nIt turns out that every infinite sum that converges in the usual way also converges to the same sum via Cesaro summability. But there are infinite sums with Cesaro sums that don't converge in the usual manner.\n\nFor example,\nsum (-1)^(n+1).\nThe partial sums look like\ns_1=1, s_2=0, s_3=1, s_4=0,....\nThis sequence of partial sums does not converge.\nHowever,\nC_1=1, C_2=1/2, C_3=2/3, C_4=1/2,...\nand it is easy to see that C_n has limit 1/2,\nso the Cesaro sum of this series is 1/2.\n\nThere are many other summability methods, for example Abel summability (which uses power series) and generalizations of Cesaro summability.\n\nIMPORTANT NOTE: the previous answerer defined c_n as the average of the original terms. THIS IS WRONG. You want the average of the partial sums.
871092
I guess you probably use the Epsilon definition for convergence? Then, you know:\nFor every Epsilon greater than 0, there exists an N such that n>N implies /xn - a/ < E'.\nYou want to show that this also implies:\n/xn^(1/3) - a^(1/3)/ < E.\n\nI think that you can probably do something with:\na^3 - b^3 = (a - b)(a^2 + ab + b^2)\nmaking x = a in this equation, the proof's a = b.\nThen, you can find an N so that /[xn^(1/3) - a^(1/3)][xn^2/3 + xn^(1/3)a^(1/3) + a^(2/3)]/ < E'. Perhaps this is easier to relate to the first one?\n\nGood luck!
303836
Nope,\n\n   The log function blows up to infinity.  (If you don't believe this, give me a value for y=f(x) = K * log(x).  Ha ha! I can give you a value for x such that f(x) is greater than the value you gave me! )\n\n  Your pal from Calc I\n   cuntsucker
199335
If you write ln[n/(n+1)]=ln n -ln(n+1), you can find the nth partial sum by writing it out and cancelling *a lot*. What is left over will be ln(1)-ln(n+1). As n->infty, this diverges,\nso the original sum diverges.
1081409
I assume you mean:\nlim n->infinity of (1+n)^(1/n)\nIn this case, the limit is one.\n\nHowever, if you do:\nlim n->infinity of (1+(1/n))^n\nit is equal to e, or 2.718.....\n\n:)
1000878
First let I(k,n)=int (log x)^k x^n dx where the integral goes from 0 to 1. Do an integration by parts with u=(log x)^k x^(n-1) and dv=x dx\nto get that I(k,n)=-k/(n+1). By induction, we then obtain that\nI(n,n)=(-1)^n n!/(n+1)^n I(0,n)\n=(-1)^n n!/(n+1)^(n+1).\n\nNow, write x^(ax)=exp(ax log x)=\nsum 1/n! (ax log x)^n=\nsum (1/n!) a^n x^n (log x)^n.\n\nNow integrate. Since the sum converges uniformly, the interchange of the sum and integral gives\nint x^(ax) dx=sum (1/n!) a^n int x^n (log x)^n dx\n= sum (1/n!) a^n I(n,n)\n=sum (-1)^n a^n /(n+1)^(n+1)\nas desired.
451479
hav u ever heard of the Principle of Mthematical Induction ?\nthat could solve ur problem in a jiffy!!\n\nHowever if u havn't, dont worry!!\n\nNote that, the sum in ur series(i.e. 1+1/2+1/3+...+1/n),where n>=2, is always less than 2, but greater than 1. so it can't b an integer.\nwanna know why?\ntry cramming up some theorems about Sequence & series of real numbers!! its a regulation part of any under-graduate level Real analysis course.