Is the sum of 1/n to infinity as n = 1 thread closed for moderation?

[nirky]

• ∞
  
• ∑ (1/n)
• n=1

The sum of 1/n to infinity as n =1

 

[jbstemp]

The series is divergent.

 

[hilbert2]

That's called harmonic series and it is divergent as jbstemp said. But on the other hand, ##\sum_{n=1}^{\infty}\frac{1}{n^{k}}## converges if ##k## is any number larger than 1.

 

[nirky]

The series is divergent.

Thanks jbstemp.

 

[nirky]

That's called harmonic series and it is divergent as jbstemp said. But on the other hand, ##\sum_{n=1}^{\infty}\frac{1}{n^{k}}## converges if ##k## is any number larger than 1.

Thanks hilbert2 but could you explain the question I asked in more details?

 

[nirky]

Thanks hilbert2 but could you explain the question I asked in more details?

Maybe tell the steps to it and does it have any awnser?

 

[HallsofIvy]

What kind of answer do you want? You have already been told the series is "divergent" which means that the infinite sum does not have any finite answer. You could prove that using the "integral test". We can extend the sum to all x rather than just integer n by using the continuous variable "x" rather than just the integer "n".
Looking at a graph of y= 1/x, you can see that the curve, between x= n and x= n+ 1, lies below the horizontal line 1/n. That tells is that the area under the curve, the integral, is less than the sum of the area of rectangles having width 1 and height 1/n, the sum of 1/n.

That area is given by [itex]\int_1^\infty \frac{1}{x} dx= \left[ln(x)\right]_1^\infty[/itex] but that improper integral has no answer since ln(x) has no limit as x goes to infinity. There is no finite "area under the curve" so there can be no finite area of the rectangles which would be larger.

 

[mathman]

Another way of seeing divergence.
1+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+1/16)+...>1+1/2+1/2+1/2+1/2+...

 

[berkeman]

Thread closed for Moderation...