91Ó°ÊÓ

The sequence $\left\{a_n\right\}=\left\{\frac{1}{n}\right\}$is a convergent sequence and converges to $0$.

The series $\underset{k=1}{\overset{∞}{∑}}\frac{1}{k}$ is a harmonic series and by p-series test the series $\underset{k=1}{\overset{∞}{∑}}\frac{1}{k}$is divergent.

Therefore, if the sequence $\left\{a_n\right\}$converges, then $\underset{k=1}{\overset{∞}{∑}}{a}_n$converges is not true.

Hence, the given statement is false.

The series $\underset{k=1}{\overset{∞}{∑}}\frac{1}{k}$ is a harmonic series and by p-series test the series $\underset{k=1}{\overset{∞}{∑}}\frac{1}{k}$is divergent.

Consider the sequence $\left\{S_n\right\}=\left\{\frac{1}{n}\right\}$.

The sequence $\left\{a_n\right\}=\left\{\frac{1}{n}\right\}$is a convergent sequence and converges to $0$.

Therefore, if the series $\underset{k=1}{\overset{∞}{∑}}{a}_k$diverges, then $\left\{S_n\right\}$is its sequence of partial sums, then $\underset{n→∞}{\lim }{S}_n=∞$is not true.

Hence, the given statement is false.

Consider the geometric series,$\underset{k=0}{\overset{∞}{∑}}{a}_k=\underset{k=0}{\overset{∞}{∑}}1$.

The series $\underset{k=0}{\overset{∞}{∑}}1$is a geometric series with common ratio $r=1$, which is equal to $1$.

The geometric series with ratio equal to $1$is divergent.

Therefore, $\underset{k=0}{\overset{∞}{∑}}{a}_k=\underset{k=0}{\overset{∞}{∑}}1$is divergent.

Again, consider the geometric series, $\underset{k=0}{\overset{∞}{∑}}{b}_k=\underset{k=0}{\overset{∞}{∑}}\left(-1\right)$.

The series $\underset{k=0}{\overset{∞}{∑}}{b}_k=\underset{k=0}{\overset{∞}{∑}}\left(-1\right)$is a geometric series with common ratio $r=1$, which is equal to $1$.

The geometric series with ratio equal to $1$is divergent.

Therefore, $\underset{k=0}{\overset{∞}{∑}}{b}_k=\underset{k=0}{\overset{∞}{∑}}\left(-1\right)$is divergent.

Consider the series,$\underset{k=0}{\overset{∞}{∑}}\left(a_k+b_k\right)$.

$$\begin{gathered} \underset{k=0}{\overset{∞}{∑}}\left(a_k+b_k\right)=\underset{k=0}{\overset{∞}{∑}}\left(1+\left(-1\right)\right) \\ =\underset{k=0}{\overset{∞}{∑}}0 \\ =0 \end{gathered}$$

The partial sum of series $\underset{k=0}{\overset{∞}{∑}}0$is a constant and hence, it is convergent.

Therefore, localid="1649657986051" $\underset{k=0}{\overset{∞}{∑}}\left(a_k+b_k\right)=\underset{k=0}{\overset{∞}{∑}}0$is convergent.

Hence, the given statement is false.

The series $\underset{k=1}{\overset{∞}{∑}}{a}_k,\underset{k=1}{\overset{∞}{∑}}{b}_k$are convergent.

Consider the series $\underset{k=1}{\overset{∞}{∑}}\left(c{a}_k+d{b}_k\right)$. Then,

$$\begin{gathered} \underset{k=1}{\overset{∞}{∑}}\left(c{a}_k+d{b}_k\right)=\left(c{a}_1+d{b}_2\right)+\left(c{a}_2+d{b}_2\right)+\left(c{a}_3+d{b}_3\right)+... \\ =\left(c{a}_1+c{a}_2+c{a}_3+...\right)+\left(d{b}_1+d{b}_2+d{b}_3+...\right) \\ =c\left(a_1+a_2+a_3+...\right)+d\left(b_1+b_2+b_3+...\right) \\ =\underset{k=1}{\overset{∞}{c∑}}{a}_k+d\underset{k=1}{\overset{∞}{∑}}{b}_k \end{gathered}$$

Hence, the given statement is true.

As it is known the adding and deleting the terms from the series does not affect the convergence of the series.

Thus, if the series $\underset{k=0}{\overset{∞}{∑}}{a}_{n+k}$converges to $5$, then the series $\underset{k=0}{\overset{∞}{∑}}{a}_{n+k}$converges to a value $L<5$is not true.

Hence, the given statement is false.

As it is known the adding and deleting the terms from the series does not affect the convergence of the series.

Thus, if the series $\underset{k=N}{\overset{∞}{∑}}{a}_k$converges, then $\underset{k=N}{\overset{∞}{∑}}{a}_k=\underset{k=0}{\overset{∞}{∑}}{a}_{n+k}$is true.

Hence, the given statement is true.

If a series is convergent, then the limit of the nth term is zero.

Thus, if a geometric series $\underset{k=0}{\overset{∞}{∑}}c{r}^k$converges, then $\underset{k→∞}{\lim }c{r}^k=0$is true.

Hence, the given statement is true.

The terms of the series $\underset{k=0}{\overset{∞}{∑}}{a}_k$are given below,

role="math" localid="1649658885403" $\underset{k=0}{\overset{∞}{∑}}{a}_k\mathit{=}\mathit{4}\mathit{+}\frac{\mathit{4}}{\mathit{2}}\mathit{+}\frac{\mathit{4}}{{\mathit{2}}^{\mathit{2}}}\mathit{+}\mathit{.}\mathit{.}\mathit{.}$

The sum of the series is given below,

$$\begin{gathered} S_{∞}=\frac{4}{1-{\displaystyle \frac{1}{2}}} \\ =8 \end{gathered}$$

Thus, if the series $\underset{k=1}{\overset{∞}{∑}}{a}_k$where $$\begin{gathered} a_1=4, \\ {a}_{k+1}=\frac{a_k}{2} \end{gathered}$$for $k>1$, converges to $7$is not true.

Hence, the given statement is false.