Difference between convergence of partial sum and series

Teachme
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I am a little confused as to notation for convergence. I included a picture too.
If you take a look it says "then the series Ʃan is divergent"
Does the "Ʃan" just mean the convergence as to the sum of the series, or the lim an as n→ ∞ nth term?
I believe it is the sum of the series but I just want to make sure.

As for the physical meaning of Ʃan I think that's related to the area, but what is the physical meaning of taking the lim an as n→ ∞ for the nth term?
Thanks
 

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You are correct that when one talks about the convergence of a series Ʃan one is referring to the convergence of its sum, or put more eloquently, the convergence of its sequence of partial sums.

As for its physical interpretation, I'm not aware of any canonical physical interpretation of infinite series in general. However, if the series converges, that is to say if its sequence of partial sums converges, then the bits added to the sum must become infinitesimally small, although this is not a sufficient requirement for its convergence; the standard counter-example being Ʃ1/n.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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