Question

Solve the following..​

Answer 1

Step-by-step explanation:

Given: $\\ \rm\gamma=\lim \limits _{n \rightarrow \infty}\left(\sum_{k=1}^{n} \frac{1}{k}-\ln n\right)$

To find: value of γ.

Solution:

Take RHS,

$\[ \begin{array}{l} \\ \displaystyle \rm =\lim _{n \rightarrow \infty}\left(\sum_{k=1}^{n} \frac{1}{k}-\ln n\right) \\ \\ \displaystyle \rm=\lim _{n \rightarrow \infty}\left(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\ln n\right) \\ \\ \displaystyle \rm=\lim _{n \rightarrow \infty}\left[\frac{1}{n}\left(\frac{n}{1}+\frac{n}{2}+\cdots+\frac{n}{n}\right)-\frac{n \ln n}{n}\right] \\ \\ \displaystyle \rm=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\left(n+\frac{n}{2}+\cdots+1\right)-n \ln n\right] \\ \\ \displaystyle \rm=\frac{1}{\infty}(\text { Non }-\text { infinite value })=0 \\ \\ \red{ \boxed{ \displaystyle \rm \therefore \text { LHS }= \gamma=0 }}\end{array} \]$

Answer: Hence the value of γ=0.

Answer 2

Answer:

Step-by-step explanation:

Given: \begin{gathered} \\ \rm\gamma=\lim \limits _{n \rightarrow \infty}\left(\sum_{k=1}^{n} \frac{1}{k}-\ln n\right) \end{gathered}

γ=

n→∞

lim

(

k=1

∑

n

k

1

−lnn)

To find: value of γ.

Solution:

Take RHS,

\begin{gathered} \begin{array}{l} \\ \displaystyle \rm =\lim _{n \rightarrow \infty}\left(\sum_{k=1}^{n} \frac{1}{k}-\ln n\right) \\ \\ \displaystyle \rm=\lim _{n \rightarrow \infty}\left(\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\ln n\right) \\ \\ \displaystyle \rm=\lim _{n \rightarrow \infty}\left[\frac{1}{n}\left(\frac{n}{1}+\frac{n}{2}+\cdots+\frac{n}{n}\right)-\frac{n \ln n}{n}\right] \\ \\ \displaystyle \rm=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\left(n+\frac{n}{2}+\cdots+1\right)-n \ln n\right] \\ \\ \displaystyle \rm=\frac{1}{\infty}(\text { Non }-\text { infinite value })=0 \\ \\ \red{ \boxed{ \displaystyle \rm \therefore \text { LHS }= \gamma=0 }}\end{array} \end{gathered}

=

n→∞

lim

(

k=1

∑

n

k

1

−lnn)

=

n→∞

lim

((

1

1

+

2

1

+

3

1

+⋯+

n

1

)−lnn)

=

n→∞

lim

[

n

1

(

1

n

+

2

n

+⋯+

n

n

)−

n

nlnn

]

=

n→∞

lim

n

1

[(n+

2

n

+⋯+1)−nlnn]

=

∞

1

( Non − infinite value )=0

∴ LHS =γ=0

the value of γ=0.