"why does the value of a converging alternating series with terms that are nonincreasing in magnitude lie between any two consecutive terms of its sequence of partial sums

The objective is to state why the value of converging alternating seies with terms that are non increasing in magnitude lie between any two consecutive terms of partial sums.

Let alternating series

Sn = partial sum of the series up to  n terms

{S2k} = sequence of partial sum of even terms

{S2k+1} = sequence of partial sum of odd terms

As the magnitude of the terms in the alternating series are non-increasing in magnitude, sequence {S2k} is bounded above by S1 and sequence {S2k+1} is bounded by S2. So, l lies between S1 and S2.

In the similar war, if first two terms of the series are deleted, then l lies in between S3 and S4 and so on.

Hence, the value of converging alternating series with terms that are non-increasing in magnitude lies between any two consecutive terms of partial sums. So, the remainder Rn = S – Sn alternating sign

15

step-by-step explanation:

we are given the following expression and we are to solve it and find the product:

-13 × (-5)

solving the given expression in the standard order of solving a mathematical expression, beginning with the opening of brackets and then multiplying the terms and changing their signs.

-3 × -5 = 15 (answer is positive because when negative is multiplied with negative, it gives a positive answer)

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!