#Summation
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Hm...
Well, you can show it converges, but I don't think it has a particularly simple value.
At least, I don't see a way of obtaining it.
It's a question meant to be solved and obtain a specific value so there must be a way
I see. Hm...
try to look at smaller “base cases” (i.e., sums of 1/(n+1)-1/(n+2) and 1/(2n+1)-1/(2n+2)) and observe what happens, and also what methods you use, to evaluate the sum in question
Look as I put values on n it's not like an telescopic series but it kind of looks like expansion of ln(x) with terms multiple of 1/3 missing but the problem is that even terms have -ve on them so this idea failed
Then I thought of integration to get a approx value but I didn't reach the answer
what exactly did you integrate
The expression in sigma
Oh yeah, integration does work!
Bruh what I think I did some error on my side
Can u do complete soln ??
I want to check for my mistake
I mean, the integral is pretty easy to calculate.
yeah i see
cause you can approx with integral test
$\frac{1}{3n+1}=\sum_{k=0}^{\infty}(-1)^k(3n)^k$
temu mod | promote to mod pls

No, that doesn't work, since n can be big.
yeah
See my approach above. It's probably the intended one.
-1<n<1
Rather, -1/3 < n < 1/3.
@pine atlas
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$\sum_{n=1}^{\infty}\int_0^1\qty(x^{3n}-x^{3n-1})\mathrm{d}x=\int_0^1\frac{1-x}{1-x^3}\mathrm{d}x$
wolfqz v1.3.2
(the integral and sum are swappable in this case, by fubini's theorem)
Yes
And sorry for not closing
The reason is bcz the intgrand converges right ??
oh okay
well there's more to it
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it's absolutely convergent
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not ONLY convergent
Ohkkk