Summing an Infinite Series: Can We Prove Divergence?

Fookie
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1+(1/2)+(1/3)+(1/4)+(1/5)...+(1/n)

can sum one prove this series is divergent?



or just tell me what the expression for sum to infintiy in terms of n is?
 
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replace with a strictly smaller divergent series eg 1+1/2+1/4+1/4+1/8+1/8+1/8+1/8+1/16... or use the integral test.
 
but i thot that the integral test is not valid for series coz the graph of 1/n will not be contiuous :eek:
 
I think you ought to look up the integral test and see what it says. The integral test is a perfectly valid way to prove this result.
 
Fookie said:
1+(1/2)+(1/3)+(1/4)+(1/5)...+(1/n)

can sum one prove this series is divergent?

Forget the first term (1), and observe that:

1/2+1/3 + 1/4+1/5+1/6+1/7 + 1/8+1/9+1/10+1/11+1/12+1/13+1/14+1/15 +...>

2*1/4 + 4*1/8 + 8*1/16 +... =

1/2 + 1/2 + 1/2 +...


Is it enough ?
 
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I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.

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