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The answer provided by Piotr is mostly correct, but there is a tiny mistake.

The maximum Tsallis entropy value is actually given by the expression: $(1−N^{1−\alpha})/(\alpha-1)$.

The denominator order is reversed.

This is because this expression is obtained when dealing with a uniform probability distribution $ P ~(P = \{\frac{1}{N}, \frac{1}{N}, ..., \frac{1}{N}\})$

Considering the Tsallis entropy defined by the expression:

\begin{align} T &= \frac{1}{\alpha-1} \sum_{j=1}^{N}(P_{j} - (P_{j})^\alpha)\\ \end{align}

In this particular scenario, all instances of $ P_{j} $ will be equal to $\frac{1}{N}$. This means the sum can be reduced to $N\times(\frac{1}{N}-(\frac{1}{N})^\alpha)$. Therefore, we can rewrite this in the following way:

\begin{align} T_{\max} &= \frac{1}{α-1}\times\left \{N\times\left[\frac{1}{N}-\left(\frac{1}{N}\right)^\alpha\right]\right\}\\ T_{\max} &= \frac{1}{\alpha-1}\times (1-N^{1-\alpha})\\ T_{\max} &= \frac{1−N^{1−\alpha}}{\alpha-1}\\ \end{align}

The answer provided by Piotr is mostly correct, but there is a tiny mistake.

The maximum Tsallis entropy value is actually given by the expression: $(1−N^{1−\alpha})/(\alpha-1)$.

The denominator order is reversed.

This is because this expression is obtained when dealing with a uniform probability distribution $ P = \{\frac{1}{N}, \frac{1}{N}, ..., \frac{1}{N}\}$

Considering the Tsallis entropy defined by the expression:

\begin{align} T &= \frac{1}{\alpha-1} \sum_{j=1}^{N}(P_{j} - (P_{j})^\alpha)\\ \end{align}

In this particular scenario, all instances of $ P_{j} $ will be equal to $\frac{1}{N}$. This means the sum can be reduced to $N\times(\frac{1}{N}-(\frac{1}{N})^\alpha)$. Therefore, we can rewrite this in the following way:

\begin{align} T_{\max} &= \frac{1}{α-1}\times\left \{N\times\left[\frac{1}{N}-\left(\frac{1}{N}\right)^\alpha\right]\right\}\\ T_{\max} &= \frac{1}{\alpha-1}\times (1-N^{1-\alpha})\\ T_{\max} &= \frac{1−N^{1−\alpha}}{\alpha-1}\\ \end{align}

The answer provided by Piotr is mostly correct, but there is a tiny mistake.

The maximum Tsallis entropy value is actually given by the expression: $(1−N^{1−\alpha})/(\alpha-1)$.

The denominator order is reversed.

This is because this expression is obtained when dealing with a uniform probability distribution $ P ~(P = \{\frac{1}{N}, \frac{1}{N}, ..., \frac{1}{N}\})$

Considering the Tsallis entropy defined by the expression:

\begin{align} T &= \frac{1}{\alpha-1} \sum_{j=1}^{N}(P_{j} - (P_{j})^\alpha)\\ \end{align}

In this particular scenario, all instances of $ P_{j} $ will be equal to $\frac{1}{N}$. This means the sum can be reduced to $N\times(\frac{1}{N}-(\frac{1}{N})^\alpha$. Therefore, we can rewrite this in the following way:

\begin{align} T_{\max} &= \frac{1}{α-1}\times\left \{N\times\left[\frac{1}{N}-\left(\frac{1}{N}\right)^\alpha\right]\right\}\\ T_{\max} &= \frac{1}{\alpha-1}\times (1-N^{1-\alpha})\\ T_{\max} &= \frac{1−N^{1−\alpha}}{\alpha-1}\\ \end{align}

The answer provided by Piotr is mostly correct, but there is a tiny mistake.

The maximum Tsallis entropy value is actually given by the expression: $(1−N^{1−\alpha})/(\alpha-1)$.

The denominator order is reversed.

This is because this expression is obtained when dealing with a uniform probability distribution $ P ~(P = \{\frac{1}{N}, \frac{1}{N}, ..., \frac{1}{N}\})$

Considering the Tsallis entropy defined by the expression:

\begin{align} T &= \frac{1}{\alpha-1} \sum_{j=1}^{N}(P_{j} - (P_{j})^\alpha)\\ \end{align}

In this particular scenario, all instances of $ P_{j} $ will be equal to $\frac{1}{N}$. This means the sum can be reduced to $N\times(\frac{1}{N}-(\frac{1}{N})^\alpha)$. Therefore, we can rewrite this in the following way:

\begin{align} T_{\max} &= \frac{1}{α-1}\times\left \{N\times\left[\frac{1}{N}-\left(\frac{1}{N}\right)^\alpha\right]\right\}\\ T_{\max} &= \frac{1}{\alpha-1}\times (1-N^{1-\alpha})\\ T_{\max} &= \frac{1−N^{1−\alpha}}{\alpha-1}\\ \end{align}

The answer provided by Piotr is mostly correct, but there is a tiny mistake.

The maximum Tsallis entropy value is actually given by the expression: $(1−N^{1−α})/(α-1)$.

The denominator order is reversed.

This is because this expression is obtained when dealing with a uniform probability distribution P (P = {$\frac{1}{N}$, $\frac{1}{N}$, ..., $\frac{1}{N}$})

Considering the Tsallis entropy defined by the expression:

\begin{align} T &= \frac{1}{α-1} \sum_{j=1}^{N}(P_{j} - (P_{j})^α)\\ \end{align}

In this particular scenario, all instances of $ P_{j} $ will be equal to $\frac{1}{N}$. This means the sum can be reduced to $N*(\frac{1}{N}-(\frac{1}{N})^α$. Therefore, we can rewrite this in the following way:

\begin{align} T_{max} &= \frac{1}{α-1}* \{N*[\frac{1}{N}-(\frac{1}{N})^α]\}\\ T_{max} &= \frac{1}{α-1}* (1-N^{1-α})\\ T_{max} &= \frac{1−N^{1−α}}{α-1}\\ \end{align}

The answer provided by Piotr is mostly correct, but there is a tiny mistake.

The maximum Tsallis entropy value is actually given by the expression: $(1−N^{1−\alpha})/(\alpha-1)$.

The denominator order is reversed.

This is because this expression is obtained when dealing with a uniform probability distribution $ P ~(P = \{\frac{1}{N}, \frac{1}{N}, ..., \frac{1}{N}\})$

Considering the Tsallis entropy defined by the expression:

\begin{align} T &= \frac{1}{\alpha-1} \sum_{j=1}^{N}(P_{j} - (P_{j})^\alpha)\\ \end{align}

In this particular scenario, all instances of $ P_{j} $ will be equal to $\frac{1}{N}$. This means the sum can be reduced to $N\times(\frac{1}{N}-(\frac{1}{N})^\alpha$. Therefore, we can rewrite this in the following way:

\begin{align} T_{\max} &= \frac{1}{α-1}\times\left \{N\times\left[\frac{1}{N}-\left(\frac{1}{N}\right)^\alpha\right]\right\}\\ T_{\max} &= \frac{1}{\alpha-1}\times (1-N^{1-\alpha})\\ T_{\max} &= \frac{1−N^{1−\alpha}}{\alpha-1}\\ \end{align}