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Full Version: [WP 34s] Simple sum procedure
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Hello all,

Would you be kind enough as to explain how the $$\sum$$ works, especially with functions like $$\frac{1}{x}$$?

For example:

Compute the following: $$\sum_{1}^{10}\frac{1}{x}$$

Answer $$\approx$$ 2.92896825397

I tried creating a program as simple as 1/x but somehow the routine fills the stack with zeros.

Thanks
Enter this program:

Code:
001: LBL 00 002: 1/x 003: END

10.001 ∑ 00

The result is 2.92896825396854

Pauli
Hello Paul,

I thought it was as simples as $$1$$ ENTER $$10$$ $$\sum$$ $$00$$. That is probably why the routine was filling the stack with zeros and returning $$+\infty$$ and an error message.

Thank you.

**EDIT: From a quick scan on the 34s user's manual, I saw that the data should be entered using the cccccc.fffii format. Alright, if that is how it is supposed to work, allow me to ask a couple of questions:

How do I separate the lower limit from the increment?
Do those 3 f's mean the number can not be greater than 999?

Marcio
(06-13-2015 12:57 PM)Thomas_Sch Wrote: [ -> ]Here is a reference:
on page 112
I didn't find the v3.1 manual's remarks in that section helpful at all, particularly because it had no example and contained an error. But the same section on page 174 of the v3.3 manual explained it very well.
(06-13-2015 08:04 PM)striegel Wrote: [ -> ]
(06-13-2015 12:57 PM)Thomas_Sch Wrote: [ -> ]Here is a reference:
on page 112
I didn't find the v3.1 manual's remarks in that section helpful at all, particularly because it had no example and contained an error. But the same section on page 174 of the v3.3 manual explained it very well.
That's correct.
But the manual v3.3 is not available for all users without ordering, so I had choosen the manual v3.1 instead.
(06-13-2015 08:04 PM)striegel Wrote: [ -> ]I didn't find the v3.1 manual's remarks in that section helpful at all

WP 34S pocket reference

Σ lbl                  x → r
Compute a sum using the routine specified at LBL.
Initially, X contains the loop control number in the format cccccc.fffii and the sum is set to 0. Each run through the routine specified by lbl computes a summand. Then, this summand is added to the sum; the operation then decrements cccccc by ii and repeats until ccccccfff.

HTH
Thomas
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