Help in infinite series problem

In summary, the equations state that if you add a sequence of terms with the same base (1 + 1/n), then the sum is 1/n.
  • #1
danunicamp
7
0

Homework Statement



Prove that:
[n+1 / n^2 + (n+1)^2 / n^3 + ... + (n+1)^n / n ^ (n+1) -> e-1

Homework Equations



I have been trying it for couple of days. Tried to work the terms, natural log it all, use the byniomial theory but i can´t get to the right answer.

The Attempt at a Solution



Working the terms:

n+1/n^2 = n( 1+1/n) / n^2 = (1+1/n)/n
(n+1)^2/n^3 = n^2 (1+1/n) ^2 / n^3 = (1+1/n)^2/n

Ok, so at n:

(n+1)^n / n^(n+1) = n^n (1+1/n) ^ n / n ^(n+1) = (1+1/n) ^n / n

Adding the terms up:

[(1+1/n) + (1+1/n)^2 +... (1+1/n)^n] / n

Adding the sequence: ( a + ar^2 + ar^3 + ar^n = a / 1 - r)

(1/n) / [1 - (1+1/n)] = (1/n) / -1/n = -1

Second try using logarithm got to:

1 - [ ln (1+1/n)^n / ln n ] and couldn,t progress further

Thank you very much in advance for any help provided.
 
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  • #2
danunicamp said:

Homework Statement



Prove that:
[n+1 / n^2 + (n+1)^2 / n^3 + ... + (n+1)^n / n ^ (n+1) -> e-1

Homework Equations



I have been trying it for couple of days. Tried to work the terms, natural log it all, use the byniomial theory but i can´t get to the right answer.

The Attempt at a Solution



Working the terms:

n+1/n^2 = n( 1+1/n) / n^2 = (1+1/n)/n
(n+1)^2/n^3 = n^2 (1+1/n) ^2 / n^3 = (1+1/n)^2/n

Ok, so at n:

(n+1)^n / n^(n+1) = n^n (1+1/n) ^ n / n ^(n+1) = (1+1/n) ^n / n

Adding the terms up:

[(1+1/n) + (1+1/n)^2 +... (1+1/n)^n] / n
Looks OK up to here.

Adding the sequence: ( a + ar^2 + ar^3 + ar^n = a / 1 - r)
No, you're only adding finitely many terms. If we put ##a = 1 + 1/n##, you have
$$\frac{1}{n}(a + a^2 + a^3 + \cdots + a^n) = \frac{a}{n}(1 + a + a^2 + \cdots + a^{n-1}) = \frac{a}{n}\left(\frac{1 - a^n}{1-a}\right)$$
Now plug in ##a = 1 + 1/n## and simplify.
 
  • #3
Thank you very very much.

I understand it now and finally solved it.

Thank you again!
 

1. What is an infinite series?

An infinite series is a sum of an infinite number of terms. It is represented by the notation Σn=1∞ a_n, where n is the index of the terms and a_n is the nth term in the series.

2. How do I know if an infinite series converges?

In order for an infinite series to converge, the limit of its terms as n approaches infinity must equal zero. This means that as n gets larger and larger, the terms in the series get smaller and smaller, eventually approaching zero.

3. What is the difference between a convergent and a divergent infinite series?

A convergent infinite series has a finite sum, meaning that the terms in the series eventually get small enough to add up to a specific value. A divergent infinite series has an infinite sum, meaning that the terms in the series do not get small enough to add up to a specific value and the sum continues to grow without bound.

4. How do I find the sum of an infinite series?

There are various methods for finding the sum of an infinite series, depending on the type of series. Some common methods include using geometric or telescoping series, or using Taylor or Maclaurin series. It is important to understand the properties and patterns of the series in order to determine the appropriate method for finding the sum.

5. Can I use a calculator to solve an infinite series problem?

While some calculators do have the capability to solve certain types of infinite series problems, it is important to understand the concepts and techniques used in solving these problems. Relying solely on a calculator may result in incorrect answers or a lack of understanding of the underlying concepts.

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