I Index and bound shift in converting a sum into integral

[Alex_F]

Considering the below equality (or equivalency), could someone please explain how the bounds and indices are shifted?
$$\sum_{i=2}^{k}(h_i/f_{i-1})=\int_{1}^{k}(h(i)/f(i))di$$

 

[Svein]

As it stands the expression is meaningless. You can approximate an integral with a sum, but the index in the sum and the number of elements in the sum cannot be carried over in the way you present it.

 

[Alex_F]

As it stands the expression is meaningless. You can approximate an integral with a sum, but the index in the sum and the number of elements in the sum cannot be carried over in the way you present it.

Actually this is not written by me. I have seen this in a book and also couple of articles. So I suppose that is correct.

 

[Mark44]

I agree with @Svein -- without some additional context, it's difficult to understand what the parts of the equation is supposed to represent. For example, is $f$ a function of some real variable, or is it a sequence? (However, a sequence is generally considered to be a function defined on some set of integers.)

 

[Delta2]

Generally what it holds, given that the function $g(i)$ is monotonically decreasing is that $$\int_N^{M+1}g(i)di\leq\sum_{i=N}^Mg(i)\leq g(N)+\int_N^{M}g(i)di$$

So as you can see we don't have equality but a pair of inequalities that "sandwiches" the sum, we don't have index shifting ($i$ remains $i$ everywhere), but the bounds are kind of shifted.
For more details check
https://en.wikipedia.org/wiki/Integral_test_for_convergence#Proof

if g is increasing then the inequalities are reversed that is it holds:
$$\int_N^{M+1}g(i)di\geq\sum_{i=N}^Mg(i)\geq g(N)+\int_N^{M}g(i)di$$

 

[Svein]

(i remains i everywhere),

In the integrals, i must be a real or complex number (if you mean a Stieltjes integral your expression must be a bit more complicated). In the summation, i is an integer.

 

[Delta2]

In the integrals, i must be a real or complex number (if you mean a Stieltjes integral your expression must be a bit more complicated). In the summation, i is an integer.

No those are Riemann integrals. I guess the symbol i is a bit overloaded there.