Sum of Series 1/n: Is it Infinity?

In summary, the conversation discusses the convergence and divergence of the sum of 1/n for n=1 to infinity, and whether the sum of 1/nn for n=1 to infinity is also infinity. Different series, such as 1/n^2 and 1/nlog(n), are mentioned and it is noted that the sum of ak converges for a<1. The concept of mind reading is also brought up.
  • #1
Chris Miller
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I've seen the proof that the sum of 1/n for = 1 to infinity is infinity (which still blows my mind a little).

Is the sum of 1/nn for n = 1 to infinity also infinity?

i.e, 1 + 2/4 + 3/27 + 4/256+...
 
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  • #2
I don't think you mean ##n^n## where ##n=1## right?

also your series should read ##1 + 1/4 + 1/27 ... ##?
 
  • #3
After the first two elements it is smaller than the sum over ##\displaystyle \frac{1}{2^n}## where you can find the sum directly.

If you want to explore the region between divergence and convergence more, have a look at ##\displaystyle \frac{1}{n^2}## and ##\displaystyle \frac{1}{n \log(n)}## and ##\displaystyle \frac{1}{n \log(n)^2}##and ##\displaystyle \frac{1}{n \log(n) \log(log(n))}##. These will need some clever tricks to determine if they converge.
 
  • #4
oops, yes, thanks, I meant the sum of n/nn for n = 1 to infinity

So... it's < 2?
 
  • #5
##\frac{n}{n^n}##? That is equal to ##\frac{1}{n^{n-1}}##
Chris Miller said:
So... it's < 2?
It is.
 
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  • #6
Chris Miller said:
I've seen the proof that the sum of 1/n for = 1 to infinity is infinity (which still blows my mind a little).

Is the sum of 1/nn for n = 1 to infinity also infinity?

i.e, 1 + 2/4 + 3/27 + 4/256+...

the sum of ak converges for a<1 for k=1 to inf - this is just a geometric series and the sum is 1/(1-a)

so by that 1/nn converges for n=1 to inf as 1/nn becomes smaller than any constant ak
 
  • #7
mfb said:
##\frac{n}{n^n}##? That is equal to ##\frac{1}{n^{n-1}}##It is.
He meant [itex]\sum\limits_{n\to\infty} \,\, \frac{n}{n^n}[/itex].
 
  • #8
Matt Benesi said:
He meant [itex]\sum\limits_{n\to\infty} \,\, \frac{n}{n^n}[/itex].
Same thing?
 
  • #9
Chris Miller said:
Same thing?
Unless you guys are mind readers, and meant [itex]\sum\limits_{n\to\infty} \,\, \frac{n}{n^n} = \sum\limits_{n\to\infty} \,\, \frac{1}{n^{n-1}}[/itex] by what you said, [itex]\frac{n}{n^n}\, =\, \frac{1}{n^{n-1}}[/itex] is different from the sum.
 

1. What is the sum of the series 1/n?

The sum of the series 1/n is known as the harmonic series and it is an infinite series that diverges, meaning it does not have a finite sum. This means that as you add more and more terms, the sum will continue to get larger and larger without ever reaching a specific value.

2. Is the sum of the series 1/n equal to infinity?

Yes, the sum of the series 1/n is equal to infinity. This can be proven mathematically using the limit comparison test or the integral test. Both methods show that the series diverges and has no finite sum.

3. Why does the sum of the series 1/n diverge?

The sum of the series 1/n diverges because the terms in the series do not approach zero as n approaches infinity. In fact, the terms in the series get larger and larger as n increases, causing the sum to also increase without bound.

4. Can the sum of the series 1/n be approximated?

Yes, the sum of the series 1/n can be approximated using different methods such as the partial sum or the Euler-Maclaurin formula. However, these approximations will always be finite and will never reach the actual sum of infinity.

5. Are there any real-life applications of the sum of the series 1/n?

Yes, the sum of the series 1/n has applications in various fields such as physics, engineering, and economics. It is used to model situations where the value of a variable increases without bound, such as in radioactive decay or population growth. It is also used in the study of infinite series and their properties.

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