Shahjahanabad and Nineteenth century Delhi .
Answers
Answer:
Solution−
Consider,
\begin{gathered}\displaystyle\rm{ \frac{1}{ {1} } + \frac{1}{ {1}^{} + {2}^{} } + \frac{1}{ {1}^{} + {2}^{} + {3}^{} } + \dots} \\ \end{gathered}
1
1
+
1
+2
1
+
1
+2
+3
1
+…
The general term is given by
\begin{gathered}\rm \: T_{n} = \dfrac{1}{1 + 2 + 3 + - - - + n} \\ \end{gathered}
T
n
=
1+2+3+−−−+n
1
\begin{gathered}\rm \: T_{n} = \dfrac{1}{\dfrac{n(n + 1)}{2} } \\ \end{gathered}
T
n
=
2
n(n+1)
1
\begin{gathered}\rm \: T_{n} = \dfrac{2}{n(n + 1)} \\ \end{gathered}
T
n
=
n(n+1)
2
\begin{gathered}\rm \: T_{n} = 2\bigg(\dfrac{n + 1 - n}{n(n + 1)}\bigg) \\ \end{gathered}
T
n
=2(
n(n+1)
n+1−n
)
\begin{gathered}\rm \: T_{n} = 2\bigg({\dfrac{1}{n} - \dfrac{1}{n + 1} }\bigg) \\ \end{gathered}
T
n
=2(
n
1
−
n+1
1
)
So,
\begin{gathered}\rm \: T_{1} = 2\bigg({1 - \dfrac{1}{2} }\bigg) \\ \end{gathered}
T
1
=2(1−
2
1
)
\begin{gathered}\rm \: T_{2} = 2\bigg({ \frac{1}{2} - \dfrac{1}{3} }\bigg) \\ \end{gathered}
T
2
=2(
2
1
−
3
1
)
\begin{gathered}\rm \: T_{3} = 2\bigg({ \frac{1}{3} - \dfrac{1}{4} }\bigg) \\ \end{gathered}
T
3
=2(
3
1
−
4
1
)
.
.
.
.
\begin{gathered}\rm \: T_{n} = 2\bigg({ \frac{1}{n} - \dfrac{1}{n + 1} }\bigg) \\ \end{gathered}
T
n
=2(
n
1
−
n+1
1
)
.
.
.
.
So,
\begin{gathered}\rm \: S_n = T_{1} + T_{2} + T_{3} + - - - + T_{n} \\ \end{gathered}
S
n
=T
1
+T
2
+T
3
+−−−+T
n
\begin{gathered}\rm \: = \:2\bigg({ 1 - \dfrac{1}{2} }\bigg) + 2\bigg({ \dfrac{1}{2} - \dfrac{1}{3} }\bigg) + 2\bigg({ \dfrac{1}{3} - \dfrac{1}{4} }\bigg) + - - + 2\bigg({ \dfrac{1}{n} - \dfrac{1}{n + 1} }\bigg) \\ \end{gathered}
=2(1−
2
1
)+2(
2
1
−
3
1
)+2(
3
1
−
4
1
)+−−+2(
n
1
−
n+1
1
)
\rm \: = \:2\bigg(1 - \dfrac{1}{2} +\dfrac{1}{2} - \dfrac{1}{3} + - - + \dfrac{1}{n} - \dfrac{1}{n + 1} \bigg)=2(1−
2
1
+
2
1
−
3
1
+−−+
n
1
−
n+1
1
)
\rm \: = \:2\bigg(1 - \dfrac{1}{n + 1} \bigg)=2(1−
n+1
1
)
\begin{gathered}\rm\implies \:\rm \:S_n = \:2\bigg(1 - \dfrac{1}{n + 1} \bigg) \\ \end{gathered}
⟹S
n
=2(1−
n+1
1
)
So, Required sum is
\begin{gathered}\rm \: \displaystyle\lim_{n \to \infty }\rm S_n \\ \end{gathered}
n→∞
lim
S
n
\begin{gathered}\rm \: =\displaystyle\lim_{n \to \infty }\rm \:2\bigg(1 - \dfrac{1}{n + 1} \bigg) \\ \end{gathered}
=
n→∞
lim
2(1−
n+1
1
)
\begin{gathered}\rm \: = \:2 \times (1 - 0) \\ \end{gathered}
=2×(1−0)
\begin{gathered}\rm \: = \:2 \\ \end{gathered}
=2
Hence,
\begin{gathered} \\ \rm\implies \:\boxed{\sf{ \:\displaystyle\rm { \frac{1}{ {1} } + \frac{1}{ {1}^{} + {2}^{} } + \frac{1}{ {1}^{} + {2}^{} + {3}^{} } + \dots = 2} \: }} \\ \\ \end{gathered}
⟹
1
1
+
1
+2
1
+
1
+2
+3
1
+⋯=2
Answer:
Old Delhi or Purani Dilli is an area in the city of Delhi, India. It was founded as a walled city named Shahjahanabad in 1639, when Shah Jahan (the Mughal emperor at the time) decided to shift the Mughal capital from Agra.