= Solution
The <coefficients> of the formal series must be <smooth functions> $H_n\in C^\infty(M)$, rather than points of $M$. Interpret complex parameter values in the complexified <function> algebra. The <Casimir function> condition for the <Poisson pencil> is a coefficientwise formal identity:
$$
0=\{H,f\}_\lambda
=\{H_0,f\}_1+\sum_{n\geq1}\lambda^n
\left(\{H_n,f\}_1+\{H_{n-1},f\}_2\right).
$$
Equality of the formal <coefficients> proves
$$
\boxed{\{H_0,f\}_1=0,\qquad
\{H_n,f\}_1=-\{H_{n-1},f\}_2\quad(n\geq1).}
$$
These are the <Lenard-Magri recursion> relations, valid for every <smooth function> $f$.
For $i\geq1$ and $j\geq0$, use the recursion in the first argument, then antisymmetry and the recursion in the second:
$$
\{H_i,H_j\}_1=-\{H_{i-1},H_j\}_2
=\{H_{i-1},H_{j+1}\}_1.
$$
Each step lowers the first index while raising the second. Iterating $i$ times gives
$$
\{H_i,H_j\}_1=\{H_0,H_{i+j}\}_1=0,
$$
because $H_0$ is a <Casimir function> of the first <Poisson structure>. This includes $i=0$ directly. The recursion one step higher then gives
$$
\{H_i,H_j\}_2=-\{H_{i+1},H_j\}_1=0.
$$
Thus the <commuting coefficients of a Poisson pencil Casimir> satisfy
$$
\boxed{\{H_i,H_j\}_1=\{H_i,H_j\}_2=0\qquad(i,j\geq0).}
$$
They are therefore <Poisson-commuting functions> for every member of the <Poisson pencil>. The argument proves involution, not <functional independence>; an infinite sequence can contain repetitions or zero <coefficients>.
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