Week -5 | Sequences & Series – 2

1.1 Logarithms of Progressions and basic properties

We define the natural logarithm of a postive number $a$ as $\log_e a = \ln a$. Logarithm of a negative number is not defined.

Let’s recollect the basic properties of natural logarithms:

• $\ln ab = \ln a+ \ln b$
• $\ln \frac{a}{b} = \ln a – \ln b$
• $\ln a^x=x\ln a$
• $\ln ab = \ln a+\ln b$
• $\ln \frac{a}{b} = \ln a – \ln b$
• $\log _b a = \log _ca. \log _bc$
• $a^{\log _ax}=x$
• $\log _{a^k}(x)=\frac{1}{k}\log_a (x)$
• $\ln 1=0$

Ex -1

If $a>1,b>1,c>1$ be 3 reals in GP, show that $\frac{1}{1+\ln a}, \frac{1}{1+ \ln b}, \frac{1}{1+ \ln c}$ are in HP

Solution:

We have $b^2=ac\thinspace \Rightarrow \ln b = \frac{\ln a+\ln b}{2}$, which implies $\ln a, \ln b, \ln c$ are in AP. This also confirms that $1+\ln a, 1+\ln b, 1+\ln c$ are in AP, and the conclusion that the required terms are in HP follows from this.

 

2.1 Sum of special sequences

Consider the sequence $a_n$ defined by $a_i=i^2$. How do we find the sum of n terms of this sequence?

What about the sum of the sequence $a_n$ defined by $a_i=i$? This seuence is in AP, and the sum of the first $n$ terms = $\frac{n(n+1)}{2}$, which should be obvious by now from what we have studied in our earlier module.

To find the sum of the squares of the first $n$ natural numbers, we observe the following:

$n^3-(n-1)^3=3n^2-3n+1$. If we sum upto n terms either sides of the equality sign, we have

$n^3=3(1^2+2^2+….+n^2)-3(1+2+3+….+n)+n$. Let’s call $1^2+2^2+…+n^2=S$, so

$3S=n^3+\frac{3n(n+1)}{2}-n\thinspace \Rightarrow S=\frac{n(n+1)(2n+1)}{6}$.

Let’s find the sum of cubes of the first n natural numbers in the same way:

We observe that $n^4-(n-1)^4=4n^3-6n^2+4n-1$. Again summing upto n terms on either sides, we have $n^4=4(1^3+2^3+…+n^3)-6(1^2+2^2+…+n^2)+4(1+2+3…+n)-n$. Calling $1^3+2^3+…+n^3=S$, we have $4S=n^4+n+6(1^2+2^2+…+n^2)-4(1+2+…+n)$. Putting in the expression for the sum of squares and sum of first n positive integers we have $S=\left \{ \frac{n(n+1)}{2} \right \}^2$.

Ex -2

Find the sum of the series : $\frac{1^2}{1}+\frac{1^2+2^2}{1+2}+\frac{1^2+2^2+3^2}{1+2+3}+…+\frac{1^2+2^2+…+n^2}{1+2+…+n}$.

Solution:

General term of this sequence : $T_r=\frac{1^2+2^2+…+r^2}{1+2+…r}=\frac{2r(r+1)(2r+1)}{6r(r+1)}=\frac{1}{3}(2r+1)$. Thus we have

$\sum\limits_{r=1}^{n}T_r=\frac{2}{3}\left(\sum\limits_{r=1}^{n}r \right )+\frac{n}{3}$.

We thus have the sum as $\frac{n(n+1)}{3}+\frac{n}{3}=\frac{n(n+2)}{3}$.

Ex-3

Find the sum upto n terms of the series $1.2.3+2.3.4+3.4.5+…$

Solution:

General term of this sequence : $T_k=k(k+1)(k+2)=k^3+3k^2+2k$.

Thus $\sum\limits_{k=1}^nT_k=\sum\limits_{k=1}^nk^3+3\sum\limits_{k=1}^nk^2+2\sum\limits_{k=1}^nk$. Plugging in the expressions for the sum of cubes, squares and first n natural numbers we have the sum as

Upon simplification this turns out to