S(n) = (1/1 + 1/2 + 1/3 + 1/4 + … + 1/n)

Question

1.33/5 (3 votes)

See more:

The goal of Part 2 is to develop an algorithm and implement code to calculate the series sum of reciprocals S(n) = (1/1 + 1/2 + 1/3 + 1/4 + … + 1/n) such that your program finds the smallest number n for which the series sum of reciprocals S(n) EQUALS OR EXCEEDS a value entered by the user.

For example, if the user enters 1.5 your program should display the following:

The smallest series sum >= 1.5 is S(2) which has the value 1.5.

What I have tried:

I have tried everything I know and can't seem to get the output to be correct.

Posted 11-Feb-18 15:56pm

Member 13673284

Updated 11-Feb-18 17:43pm

Add a Solution

Comments

[no name] 11-Feb-18 22:09pm

Show us your code. Don't be shy.

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Andy Allinger · Answer 1 · 2018-02-11T17:43:00

The harmonic series has a lot of roundoff error for large N. Does your program work for N = 10?
For N = 1,000,000 S = 14.393, and the difference between adjacent terms may be less than roundoff.
Use compensated addition Kahan summation algorithm - Wikipedia[^]
and store each term in an array S[]. Then binary search the array for the target value.