106ans is:
Gm/1 + Gm/4 + Gm/9 + ......
= Gm(1+1/4+1/9+1/16+........)
15 Answers
106
1asish r u telling it will be Gm(∞).....lolzzzz.......[1][1][1][1][1][1].......
1that's wat i'm asking...hw 2 solve the series (1+1/4+1/9+1/16+........)??
and it won't b ∞...as its a decreasing series..
1no it will tend to infinity shreyan.......(1+1/4+1/25.....)---->∞....i think so!!!!...[1][1][1][1][1]........
106no terminator the terms get smaller so it wont tend to infinity
24converging series hai yaar ..........to sum infinity kaise hoga???
1d asnwer is
lim(x→1)∫(-1)log(1-x)/x
along wid othr terms like Gm n all
1ok..so....then i think it will tend to 1 becaz 1>>1/4>>1/25>>........so i think rest of the terms r neglegible correct me if im wrong.......[1][1][1][1][1]....
1so i think the answer is Gm........i think so.......[1][1][1][1][1].......
1@integrations, ...gr8!
hw did u get this ans?? [lim(x→1)∫(-1)log(1-x)/x]
can u please post the solution???
1:P
-log(1-x)=x+x^2/2+x^3/3........ ∞
dividin thruout by x
we get
-log(1-x)/x =1+x/2+x^2/3.....∞
nw integratin it we get
d required resuts :)
66The series
1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\ldots \textrm{ to }\infty=\dfrac{\pi^2}{6}