MHB Sum of Series $\approx$ Error Estimation

Click For Summary
The series $\sum_{n=1}^{\infty}\frac{1}{3^n+4^n}$ can be approximated by summing the first 10 terms. As $n$ increases, the term $3^n$ becomes negligible compared to $4^n$, allowing for the approximation $1/(3^n+4^n) \approx 1/4^n$. The upper bound for the truncation error is given by the integral $\int_{10}^{\infty} 1/4^x\; dx$. A lower bound for the error is determined by the first neglected term, which is $1/(3^{11}+4^{11})$. This analysis provides a clear estimation of the series sum and its associated error.
ineedhelpnow
Messages
649
Reaction score
0
use the sum of the first 10 terms to approximate the sum of the series. Estimate the error.

$\sum_{n=1}^{\infty}\frac{1}{3^n+4^n}$
 
Physics news on Phys.org
ineedhelpnow said:
use the sum of the first 10 terms to approximate the sum of the series. Estimate the error.

$\sum_{n=1}^{\infty}\frac{1}{3^n+4^n}$

I will leave you to sum the first ten terms, then we observe:

As $n$ becomes large $3^n \ll 4^n$ so $1/(3^n+4^4)\approx 1/4^n$ and $1/(3^n+4^4) < 1/4^n$ so the remainder after summing the first 10 terms of the series is approximately equal to (and less than) $\int_{10}^{\infty} 1/4^x\; dx$ ...

The above is an upper bound on the truncation error, in this case a lower bound is provided by the first neglected term: $1/(3^{11}+4^{11})$

.
 
Last edited:

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K