# How can the value of an infinite series be a number having an exact value(and not a number which approaches some exact value)?

I gotta a following quetion for you

How can functions and their respective expansions(infinite series) be equal in their respective value since value of a function(for its domain as input) will be an exact number but infinite series(expansion) gives answer which is an approaching value and not an exact number?

Let me first write some functions and their respective expansions

$$(1+x)^n=1+nx+ \frac{n(n-1)x^2}{2!}+ \frac{n(n-1)(n-2)x^3}{3!}+.......∞$$ $$\tag*{ for \space |x|<1, \space n∈Q \space.....[1] }$$ $$sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}.......∞$$ $$\tag*{ x∈R \space .....[2] }$$ $$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}.......∞$$ $$\tag*{ x∈R \space .....[3] }$$ Now let me take them one by one.

If we subsitute for $$x=\frac{1}{2}$$ in $$[1]$$ we get the following

$$\frac{4}{9}=\frac{3}{4}+\frac{1}{2}+\frac{5}{16}.......∞$$

Now the L.H.S(left ahnd side) is an exact value i.e. $$\frac{4}{9}$$ but since R.H.S(right hand side) is an infinite series it can't be an exact value it must be a value that approaches some exact value but never equal to that exact value. If this is so then how come equality holds?

Now if I subsitute for $$x=\frac{π}{6}$$ in $$[2]$$ we get the following

$$\frac{1}{2}=\frac{π}{6}-\frac{π^3}{1296}+\frac{π^5}{933120}.......∞$$

On L.H.S we have an exact value i.e. $$\frac{1}{2}$$ but on R.H.S we have an infinte series whose answer will be again an approaching value but not an exact value then how come equality holds?

Similarly if we subsitute fow $$x=1$$ in $$[3]$$ we get $$e$$ on the L.H.S which is an exact value but on R.H.S we get an infinite series whose answer again should approach some value but not equal to it. So how come equality holds.

In the exact same way I can go on for the other functions and their expansions. Everytime the function will give an answer which will be an exact value(for it's domain) but its respective exapansion will be always be an infinite series and therefore will give an answer which will always be an approaching value and therefore proving the equality to be wrong.

NOTE1: It is possible that I may be gettimg somewhere wrong in my understanding. You are requested to please point out to me my mistake(so that I can digest my question) rather than opposing it.

NOTE2: Please tag my question if you find some other appropriate tag also(as I could only find two)

• Same as with any other limit. The value of an infinite series is defined to be the limit of the sequence of its partial sums. – lulu Apr 2 '20 at 14:15

Your question rests on a fundamental misunderstanding of series.

It is not true that ‘infinite series(expansion) gives answer which is an approaching value and not an exact number’: a convergent series represents a single specific number. For example,

$$\sum_{n\ge 0}\frac1{2^n}=\frac11+\frac12+\frac14+\frac18+\ldots$$

is equal to $$2$$ every bit as much as $$1+1$$ is equal to $$2$$.

Let $$\sum_{k\ge 1}a_k$$ be a series. For each $$n\ge 1$$ let $$s_n=\sum_{k=1}^na_k$$ be the partial sum of the first $$n$$ terms. These partial sums are just ordinary sums of finitely many numbers. We now consider the sequence, $$\langle s_n:n\ge 1\rangle$$, of these partial sums. If it diverges, we say that the series is divergent and don’t try to evaluate it. If, however, the sequence of partial sums converges to some limit $$L$$, we define $$\sum_{k\ge 1}a_k$$ to be that limit $$L$$. $$L$$ is a specific number, and therefore so is $$\sum_{k\ge 1}a_k$$. It is the partial sums that approach but are (usually) not equal to $$L$$; the series itself is by definition exactly $$L$$.

• @ Brian M. Scott does that mean whenever we will be given some infinite series to be calculated we are always supposed to calculate the limit of the converging value ? – dRIFT sPEED Apr 2 '20 at 14:43
• @pRSmHJN1: Yes, if you’re asked to find the sum of an infinite series, what’s wanted is the limit of the sequence of partial sums of that series, assuming that the limit exists. – Brian M. Scott Apr 2 '20 at 14:47
• Thank you very much for removing my fundamental misunderstanding of infinite series – dRIFT sPEED Apr 3 '20 at 8:22
• @pRSmHJN1: You're very welcome. – Brian M. Scott Apr 3 '20 at 14:08