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### Limit of a Series

```Date: 03/02/2003 at 13:39:58
From: Hilaf Hasson
Subject: A Limit Problem

What is the limit of the following series:
a(n) = ((1^n) + (2^n) + ... + (1998^n))^(1/n)

```

```
Date: 04/01/2003 at 23:05:51
From: Doctor Nitrogen
Subject: Re: A Limit Problem

Hi, Hilaf:

a(n) = ((1^n)+(2^n)+...+(1998^n))^(1/n)

you can first compute the value of

(A)   (1^n)+(2^n)+...+(1998^n)

by using something called Bernouilli Polynomials B_n(x). That's to be
read 'B subscript n of x', or B(x) but with the 'B' subscripted with
n. Then you will be able to compute

a(n) = ((1^n)+(2^n)+...+(1998^n))^(1/n)

for any positive integer value of n.

Let kCi denote the combinatorial symbol. Bernouilli polynomials are
found from

SUM(i = 0 to k)kCi(B_k-i times x^i).

The first four Bernouilli polynomials are:

1. B_0(x)= 1,

2. B_1(x)= x - 1/2,

3. B_2(x) = x^2 - x + 1/6,

4. B_3(x) = x^3 - (3/2)x^2 + x/2, etc.

The coefficients appearing in the polynomials are called Bernouilli
numbers and they are given by computing

SUM(i = 0 to k = k-1)kCi(B_i) = 0.

There is a formula you can use to compute (A):

SUM(x = 1 to k)x^n

= [B_n+1(k + 1) - B_n+1(0)]/n+1.

using this formula, (A) is found:

(B)   (1^n)+(2^n)+...+(1998^n)

= [B_n+1(1999) - B_n+1(0)]/n+1,

where B_n+1(1999) and B_n+1(0) are the (n+1)-st Bernouilli polynomial
with arguments

x = 1998 + 1 = 1999 and

x = 0, respectively.

You can use (B) to calculate (A), then take the n-th root (1/n) for
any positive integer value of n.

If you are interested in further research into Bernouilli polynomials
and Bernouilli numbers, there is an extraordinary paper on this topic:

"A New Approach to Bernouilli Polynomials," by D.H. Lehmer,
American Mathematical Monthly, vol. 95, #10, December, 1988, p. 905.

You can also find information on Bernouilli polynomials/numbers in:

"Handbook of Mathematical Functions," Abramowitz, Stegun, editors.

mathematics problem.

- Doctor Nitrogen, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

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