### Completely lost My task is to define a function that accepts i as as an argument to compute the following summation, return the summation from the function

m(i) 1/2 + 2/3 + 3/4 + ... + i / (i + 1), where i >= 1

halp Show us what you've tried.

You're going to need:
- a for loop to iterate a counter variable (e.g. n) from 1 to i, inclusive.
- a variable to accumulate the sum
- floating-point division, for example
`sum += static_cast<double>(n) / (n + 1);`

Tutorials:
http://cplusplus.com/doc/tutorial/control/
http://cplusplus.com/doc/tutorial/functions/
Last edited on @OP
Here's a supplementary start, with an alternative hack to the static_cast - try it with 1 instead of 1
Your next challenge is to design the function. Perhaps even get user input for the limit and produce some user-friendly output.

 ``123456789101112131415`` ``````#include int main() { int LIMIT{3}; double sum{0}; for(int i = 1; i < LIMIT; i++) { sum += i / (1. * ( i + 1 )); } std::cout << sum << '\n'; return 0; }``````

 ```1.16667 Program ended with exit code: 0 ``` ``123456`` ``````#include double sum( int i ){ return i ? i / ( i + 1.0 ) + sum( i - 1 ) : 0; } int main() { std::cout << sum( 10 ) << '\n'; }`````` this is an interesting series. it appears to approach N, always slightly under it. ```lim𝑛→∞ 𝑛/𝑛+1 = lim 𝑛→∞ 𝑛⋅1/𝑛 /( (𝑛+1)⋅1/𝑛 ) = lim 𝑛→∞ 1/(1+1/𝑛) = 1```
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