SERIES AND LIMITS
Sum the series
∑Nn=1(n+1)(n+3).Thenth term in this series is
un=(n+1)(n+3)=n^2 +4n+3,and therefore we can write
∑Nn=1(n+1)(n+3)=∑N
n=1(n^2 +4n+3)=
∑N
n=1n^2 +4∑N
n=1n+∑N
n=13
=^16 N(N+ 1)(2N+1)+4×^12 N(N+1)+3N
=^16 N(2N^2 +15N+ 31).
4.2.6 Transformation of seriesA complicated series may sometimes be summed by transforming it into a
familiar series for which we already know the sum, perhaps a geometric series
or the Maclaurin expansion of a simple function (see subsection 4.6.3). Various
techniques are useful, and deciding which one to use in any given case is a matter
of experience. We now discuss a few of the more common methods.
The differentiation or integration of a series is often useful in transforming anapparently intractable series into a more familiar one. If we wish to differentiate
or integrate a series that already depends on some variable then we may do so
in a straightforward manner.
Sum the seriesS(x)=x^4
3(0!)+
x^5
4(1!)+
x^6
5(2!)+···.
Dividing both sides byxwe obtain
S(x)
x=
x^3
3(0!)+
x^4
4(1!)+
x^5
5(2!)+···,
which is easily differentiated to give
d
dx[
S(x)
x]
=
x^2
0!+
x^3
1!+
x^4
2!+
x^5
3!+···.
Recalling the Maclaurin expansion of expxgiven in subsection 4.6.3, we recognise that
the RHS is equal tox^2 expx. Having done so, we can now integrate both sides to obtain
S(x)/x=∫
x^2 expxdx.