I have the following 'homework style' question which I will quote word for word:
A test is done to check for a predicted atomic spectral line by counting the number of photons emitted from a sample in a narrow frequency range. The hypothesis under test is that no such spectral line exists. Due to background light, the detector will also see other photons and the average number of these background photons is known to be $0.85$. What is the lowest number of observed photons in the experiment that would rule out the hypothesis at the $10$% $\color{red}{\text{confidence}}$ level?
Marked in red: I am almost certain that the word 'confidence' should be 'significance'.
In the answer, it states that:
The existence of the spectral line can only cause the number of photons to be larger, so the rejection region is at large numbers of photons. For a $10$% significance level, then the rejection region is $n_r$ and above, where $$\fbox{$\color{blue}{0.1\ge \sum_{n=n_r}^{\infty}P(n; \mu=0.85) \implies 0.9 \le \sum_{n=0}^{n_r-1}P(n; \mu=0.85)}$}\tag{A}$$
The $P$ here denotes the Poisson distribution and my question is regarding how the LHS of ($\mathrm{A}$) was transformed to the RHS:
So starting from the LHS (which I understand):
$$0.1\ge \sum_{n=n_r}^{\infty}P(n; \mu=0.85)$$ and subtracting $1$ from both sides means that $$0.1-1\ge \sum_{n=n_r}^{\infty}P(n; \mu=0.85)-1$$
Now, since by definition $$\sum_{n=0}^{\infty}P(n; \mu=0.85)\equiv 1$$ I can write $$-0.9\ge \sum_{n=n_r}^{\infty}P(n; \mu=0.85)-\sum_{n=0}^{\infty}P(n; \mu=0.85)$$
Multiplying both sides by $-1$ flips the inequality sign, so
$$0.9\le \sum_{n=0}^{\infty}P(n; \mu=0.85)-\sum_{n=n_r}^{\infty}P(n; \mu=0.85)$$
But this is not the same as the RHS of ($\mathrm{A}$):
$$0.9 \le \sum_{n=0}^{n_r-1}P(n; \mu=0.85)\tag{B}$$
unless there is a way to manipulate the upper index of ($\mathrm{B}$) such that $$\sum_{n=0}^{\infty}P(n; \mu=0.85)-\sum_{n=n_r}^{\infty}P(n; \mu=0.85)=\sum_{n=0}^{n_r-1}P(n; \mu=0.85)\tag{C}$$ assuming I haven't made a mistake so far.
Could someone please help me show how to reach the RHS of ($\mathrm{A}$), or, show that ($\mathrm{C}$) is true? Any hints or tips are most appreciated, many thanks.
Let $a_n = P( n ; \mu = 0.85)$.
Then, for any $m\geq 0$, as the series $\sum_{n=0}^\infty a_n$ converges, we have $$ \sum_{n=0}^\infty a_n = \sum_{n=0}^m a_n + \sum_{n=m+1}^\infty a_n \tag{1} $$ Rearranging, this is equivalent to $$ \sum_{n=0}^\infty a_n - \sum_{n=m+1}^\infty a_n = \sum_{n=0}^m a_n \tag{2} $$ Taking $m = n_r-1$, we get $$ \sum_{n=0}^\infty a_n - \sum_{n=n_r}^\infty a_n = \sum_{n=0}^{n_r-1} a_n \tag{3} $$ This is (C).