How to check convergence of sum(cos(wn)/n) n = 1... -->infinity?
958292
I think that the sum converge for all w except w=0. Here is an idea you could try to develop.\nLets define a new serie b(n) like this: b(1) = a(1) + a(2) + ... + a(p1) such that a(1), a(2), ... a(p) have all the same sign and a(p1 + 1) the opposite sign. \nThus b(2) = a(p1 + 1) + .... a(p2) such that a(p1 + 1), ... a(p2) have the same sign and a(p2 + 1) the opposite sign.\nThus b(3) = ....  etc   Note that this construction is possible only if w is different from 0.\nThen, by construction the b serie is an alternating one. It is easy to verify that |b(n)| is a serie which converges to zero. The problem would be ended if we had |b(n+1)|<|b(n)|, but that is not true. Nonetheless, you can make an integral approach to find a majoration of the |b(n)|. [something like a majoration of |integral of cos(wx)/x between pi*n/w and pi*(n+1)/w |]\nI leave you the detail of calculation... this approach should work ...
740766
It is not possble!\n\nConsider the Fibonacci sequence 0, 1, 1, 2, 3, 5, 8 ..... This is clearly unbounded as |Un| > |Un-1|\n\n1, x, 1 + x, 1 + 2x, 2 + 3x. 3 + 5x, 5 + 8x ......\n\nIf x = 0 this is the Fibonacci sequence starting at U1 and so is still unbounded\n\nThe general term of the above sequence is therefore Un+1 + x* Un (Un, Un+1 of the Fibonacci sequence)\nNow Un diverges and so xUn also diverges\nIf x > 0, the divergence is more rapid\nif x < 0 the sequence oscillates without bound\n\nNB 1, 0, 1, 0, 1, 0, 1, ... does not follow the rules set \n\nshould go 1, 0, 1, 1. 2, 3, 5, ......
189237
The sum from n = 0 to infinity of a^n is called a geometric series. It converges if and only if |a| < 1.\n\nIn this case we have the sum from n = 1 to infinity of (x^n)/(9^n), which can also be written as (x/9)^n. So in this case a = x/9, and in order to converge we need |x/9|<1. Thus the interval of convergence is -9 < x < 9.\n\nAs for the sum of the series, well the sum from n = 1 to infinity of a^n is a/(1-a). Here a = x/9, so you've got (x/9)/(1-x/9) = x/(9-x).\n\nBTW, the series is still geometric even if x is not a natural number. Also x needs to be greater than -9.
50930
I know it.\nIts one more than the Infinity.\nIf you know the value of infinity then substitute it and you'll get the answer.
1136716
2r<22\nr<11\nAnswer= (-infinity,11)\n\n* don't have the infinity symbol on computer
887960
You're answer is perfectly correct.  You do want to use root test, and you applied it properly.\n\nYou could also check your work by sorta "thinking" it out.\n\nthe first term (if you start with n=1) is 1, then 1/4 then 1/27 then 1/256 and so on, so as you can see, if you sum the first 3 terms we get a number close to 1.3, if we sum the first 4 numbers its still about 1.3... doesn't matter how many terms it is, i think it ends up around 1.3 or 1.33333 or something, there is a way to figure out what the sum converges to.
2367
C = A - B = infinity - infinity = undefined :-), not 0.\n\nIf you have the specific series for A and B, then you could technically substitute them into C and figure out C for different {A,B}.\n\nThe short answer is this: not all infinity series are the same, so their difference is really not zero.
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
766268
This series converges to \n(sqrt(3)*pi/4)-(4ln(2))+\n((9/4)ln(3)).
160587
In each triple, the difference between the 2nd and 3rd terms is > 0 so \n\n1 + 1/2 - 1/3 + 1/4 + 1/5 -1/6 + 1/7 > 1 + 1/4 + 1/7 + 1/10 + ...\n\nThe nth term of the series on the right is 1/(3n+1) > 1/(3n+3n) = (1/6) x (1/n)  which is going to diverge.
272208
The answer is quite simple! By using logarithmic series.\n\nFor any number x+1 , the natural log can be expressed as\n\nln(1 + x) = x - x^2 / 2  +  x^3/3  -  x^4/4  + ..... (infinity)\n\nNow you must be amazed that who the heck would calculate this infinite series! This series may seem infinite but it has a finite answer. It is analogous with summation of infinite geometric series (i hope you know this term).
105048
Directly, there is no solution.  If you use a series expansion around x=0, \nsin(x)=sum((-1)^n(x^(2n+1))/(factorial(2n+1)))\nDivide this series by x, then integrate.  After the 3rd term, the error term goes to the ten millionths digit, so I'd say you're pretty safe :)\n\nsin(x)/x = 1 - (x^2)/6 + (x^4)/120 - (x^6)/5040 + .....\nIntegrated is:  x - (x^3)/18 + (x^5)/600 - (x^7)/32580 +\nor \nsum((-1)^n)(x^(2n+1))/(2n+1)*factorial(2n+1))
200234
if this series goes upto infinity then just find the nth term which will be 2n-4.4 \nthen let E represent sigma notation\n∞\nE 2n-4.4 \nn=1\n\ni hope i was help
875583
Also, If 0<a<1, this series converges because eventually each term becomes zero.  So its sum would not be infinity.  Right?
402447
No.\nYou have to find cos(100) and then multiply it by 24\ncos(100) = -.1736
203992
a) [1,2] is a sub set of domain of g so for every x in [1,2]\nwe have a unique value of g.\n\nb) if the sequence converge, in the limit Xn+1 must be equal to X, so if the limit is L we have:\n\nL=(3L^2+3)^1/4\n\nL^4=3L^2+3\n\nL^4-3L^2-3=0\n\ndelta=9+12=21\n\nL^2=(3+√21)/2  (L^2>0)\n\nL = √[(3+√21)/2]  (L>0 because X0>0)\n\n4<√21<5  ---> 7<3+√21<8\n\n---> 3<(3+√21)/2<4  ---> 1<L<2\n\nc) we have (Xn+1)^4=3Xn^2+3 let Yn be Xn^2\n\n---> (Yn+1)^2=3Yn+3 ---> Yn
486703
Unfortunately to completely understand this you need to understand the concept of limits and convergence.\nJust see that\n.9bar =.9+.9/10+.9/100+.9/1000+.9/10000        .....\nThis forms a series .\nA series converges if partial sums tend to some limit if numbers of terms is made large.\ni,e if I extend this sum to infinite number of terms the answer is finite.\nIt turns out this is a geometric series and sum is indeed 1.\nYou can understand the method you used by just considering the .9bar as a sum.That is x is a sum.This kind of thing is allowed if series is absolutely convergent.