Summing an Infinite Series: Can We Prove Divergence?

AI Thread Summary
The discussion focuses on proving the divergence of the infinite series 1 + (1/2) + (1/3) + (1/4) + ... + (1/n). Participants suggest using the integral test, which is deemed valid for this series despite initial doubts about continuity. A comparison with a smaller divergent series is also mentioned to illustrate the divergence. The argument emphasizes that ignoring the first term and analyzing the remaining series leads to a sum that diverges. Overall, the series is confirmed to be divergent through various mathematical approaches.
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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 ?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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