Square root-Taylor expansion method

The Taylor series for  f⁢(x)=1+x  using the

is given in the table below for the first few .

Taylor Series for f⁢(x)=1+x

The general coefficient of the expansion, for n≥2
is:

Square root-Taylor expansion method

L’Hôpital’s Rule () is one analytic approach,
but in many cases, using Taylor expansion is even easier
and straightforward. Essentially, Taylor expansion approximates
complicated functions by polynomials, whose limits are easy to evaluate.
We illustrate the method below.

so that we do not have to worry about the pesky negatives any more.
Then, with the help of the binomial formula:

(“o” is Landau notation)
we obtain:

This example is admittedly artificial; it was made to be annoying to solve using
L’Hôpital’s Rule alone, but much simpler if one knows how to use
the Taylor expansions:

So we compute:

Notice that all the essential information to evaluate the limit is contained in the first few derivatives of
the functions involved at particular points — in the above example,
only at x=0.
This information can be obtained by manipulating series,
unlike L’Hôpital’s Rule
which necessitates computing the derivative functions at all points.
So even monstrous expressions like this one is tractable
with Taylor expansion:

On the other hand, there definitely are situations where
L’Hôpital’s Rule works but Taylor expansion does not: for instance,

because 1/ln⁡x cannot be expanded in a Taylor series about x=0.

Our intuition suggests that it does not, because
arctan⁡(n-1) should be approximately n-1, and ∑nn-1 diverges.
However, the standard comparison test does not work
because arctan⁡x≤x (for x≥0) has the inequality in the wrong direction.
But with Taylor expansion the solution is a snap.
By expanding

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and summing both sides, we get

As ∑nO⁢(n-3) converges (being dominated by C⁢∑nn-3 for some constant C), ∑narctan⁡(n-1) must diverge (to ∞).

(Of course, this problem could be solved by using the integral test,
but who really wants to integrate ∫arctan⁡(1/x)⁢𝑑x?)

Square root-Taylor expansion method

A known: , begging。 The question turns into,solution。

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